Work - MCAT Chemical and Physical Foundations of Biological Systems
Card 1 of 84
Two children are playing on an icy lake. Child 1 weighs 50kg, and child 2 weighs 38kg. Child 1 has a backpack that weighs 10kg, and child 2 has a backpack that weighs 5kg.
Over the course of the afternoon, they collide many times. Four collisions are described below.
Collision 1:
Child 1 starts from the top of a ramp, and after going down, reaches the lake surface while going 
and subsequently slides into a stationary child 2. They remain linked together after the collision.
Collision 2:
Child 1 and child 2 are sliding in the same direction. Child 2, moving at 
, slides into child 1, moving at 
.
Collision 3:
The two children collide while traveling in opposite directions at 
each.
Collision 4:
The two children push off from one another’s back, and begin moving in exactly opposite directions. Child 2 moves with a velocity of 
.
Consider that child 1 adds 5kg of books to her backpack. She then prepares to slide down the ramp to the lake surface. If she slides down the ramp but stops herself after she has lost 2m in vertical height, how much work does the force of gravity perform on her to get her to her new position? Ignore assume the ramp is frictionless.
Two children are playing on an icy lake. Child 1 weighs 50kg, and child 2 weighs 38kg. Child 1 has a backpack that weighs 10kg, and child 2 has a backpack that weighs 5kg.
Over the course of the afternoon, they collide many times. Four collisions are described below.
Collision 1:
Child 1 starts from the top of a ramp, and after going down, reaches the lake surface while going
Collision 2:
Child 1 and child 2 are sliding in the same direction. Child 2, moving at
Collision 3:
The two children collide while traveling in opposite directions at
Collision 4:
The two children push off from one another’s back, and begin moving in exactly opposite directions. Child 2 moves with a velocity of
Consider that child 1 adds 5kg of books to her backpack. She then prepares to slide down the ramp to the lake surface. If she slides down the ramp but stops herself after she has lost 2m in vertical height, how much work does the force of gravity perform on her to get her to her new position? Ignore assume the ramp is frictionless.
Tap to reveal answer
The amount of work performed to get child 1 down the ramp is equal to the amount of kinetic energy gained by her from top to bottom, via the work-energy theorem. Unfortunately, we are not given her kinetic energy. We can't use the 5m/s in the passage, because she stops herself before sliding to the lake surface.
We must instead recognize the potential energy lost is equal to kinetic energy gained. The potential energy that she has lost is given by mgh.
The amount of work performed to get child 1 down the ramp is equal to the amount of kinetic energy gained by her from top to bottom, via the work-energy theorem. Unfortunately, we are not given her kinetic energy. We can't use the 5m/s in the passage, because she stops herself before sliding to the lake surface.
We must instead recognize the potential energy lost is equal to kinetic energy gained. The potential energy that she has lost is given by mgh.
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Two children are playing on an icy lake. Child 1 weighs 50kg, and child 2 weighs 38kg. Child 1 has a backpack that weighs 10kg, and child 2 has a backpack that weighs 5kg.
Over the course of the afternoon, they collide many times. Four collisions are described below.
Collision 1:
Child 1 starts from the top of a ramp, and after going down, reaches the lake surface while going 
and subsequently slides into a stationary child 2. They remain linked together after the collision.
Collision 2:
Child 1 and child 2 are sliding in the same direction. Child 2, moving at 
, slides into child 1, moving at 
.
Collision 3:
The two children collide while traveling in opposite directions at 
each.
Collision 4:
The two children push off from one another’s back, and begin moving in exactly opposite directions. Child 2 moves with a velocity of 
.
Before collision 3 takes place, suppose that child 1 starts at the top of a ramp that is 3m long, and 
to the lake surface.
With what speed does child 1 hit the lake surface? Ignore friction and air resistance.
Two children are playing on an icy lake. Child 1 weighs 50kg, and child 2 weighs 38kg. Child 1 has a backpack that weighs 10kg, and child 2 has a backpack that weighs 5kg.
Over the course of the afternoon, they collide many times. Four collisions are described below.
Collision 1:
Child 1 starts from the top of a ramp, and after going down, reaches the lake surface while going
Collision 2:
Child 1 and child 2 are sliding in the same direction. Child 2, moving at
Collision 3:
The two children collide while traveling in opposite directions at
Collision 4:
The two children push off from one another’s back, and begin moving in exactly opposite directions. Child 2 moves with a velocity of
Before collision 3 takes place, suppose that child 1 starts at the top of a ramp that is 3m long, and
With what speed does child 1 hit the lake surface? Ignore friction and air resistance.
Tap to reveal answer
This is an application of the work-energy theorem. The amount of work done by the force of gravity to move child 1 down the ramp is equal to force * distance, as well as equal to the amount of kinetic energy picked up down the ramp.
The force down the ramp is equal to mg*sin(30o) = 600J * 0.5 = 300J
This is an application of the work-energy theorem. The amount of work done by the force of gravity to move child 1 down the ramp is equal to force * distance, as well as equal to the amount of kinetic energy picked up down the ramp.
The force down the ramp is equal to mg*sin(30o) = 600J * 0.5 = 300J
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What is the work done by gravity if a 50kg block is pushed along a 40m track?
What is the work done by gravity if a 50kg block is pushed along a 40m track?
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The work done by any force that is perpendicular to the displacement is equal to zero. Since the block is moving horizontally, the net force in the vertical direction will be equal to zero; therefore, work done by either gravity or the normal force will be equal to zero.
The work done by any force that is perpendicular to the displacement is equal to zero. Since the block is moving horizontally, the net force in the vertical direction will be equal to zero; therefore, work done by either gravity or the normal force will be equal to zero.
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An object of mass 100g moves in circular path of radius 0.5m, under the influence of a 15N force directed radially inwards towards the center of the path. How much work is done by this force as the object moves one quarter of the way around the circle?
An object of mass 100g moves in circular path of radius 0.5m, under the influence of a 15N force directed radially inwards towards the center of the path. How much work is done by this force as the object moves one quarter of the way around the circle?
Tap to reveal answer
Work is given by 
, the dot product of force and displacement. Since the dot product only sees the components of vectors which are parallel to each other, the dot product of two perpendicular vectors is 0. Force is directed radially inward, while displacement is directed tangent to the circumference. At any point along this object's path, the force is perpendicular to displacement, so 
is simply 0.
Work is given by
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A 2kg box slides down a ramp a distance of two meters before it reaches the ground. The ramp has an angle of 30o. The coefficient of kinetic friction for the ramp is 0.1.
What is the work done by gravity on the box?
A 2kg box slides down a ramp a distance of two meters before it reaches the ground. The ramp has an angle of 30o. The coefficient of kinetic friction for the ramp is 0.1.
What is the work done by gravity on the box?
Tap to reveal answer
Work is determined using the equation 
. Here, 
is the force applied, 
is the displacement of the object, and 
is the angle of the force relative to the movement of the object. Since gravity is acting on the box, we can solve for the force of gravity causing the movement of the box. Note that in this case 
refers to the angle between gravity and the box's path; thus, the angle will be 60o, rather than 30o.
Notice how the work done by gravity is equal to the potential energy of the box at the top of the ramp. This is because mechanical energy is conserved in the system; thus, we can set the two equations equal to each other.
Work is determined using the equation
Notice how the work done by gravity is equal to the potential energy of the box at the top of the ramp. This is because mechanical energy is conserved in the system; thus, we can set the two equations equal to each other.
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A 2kg box slides down a ramp a distance of two meters before it reaches the ground. The ramp has an angle of 30o. The coefficient of kinetic friction for the ramp is 0.1.
What is the work done by the normal force on the box?
A 2kg box slides down a ramp a distance of two meters before it reaches the ground. The ramp has an angle of 30o. The coefficient of kinetic friction for the ramp is 0.1.
What is the work done by the normal force on the box?
Tap to reveal answer
Remember that the work done by a force on an object is dependent on the angle of the force to the object's displacement. The normal force acts perpendicularly to the ramp, which means it has an angle of 90o with respect to the box's displacement.
Because 
, the total work done on the box by the normal force is 
.
Remember that the work done by a force on an object is dependent on the angle of the force to the object's displacement. The normal force acts perpendicularly to the ramp, which means it has an angle of 90o with respect to the box's displacement.
Because
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A 2kg box slides down a ramp a distance of two meters before it reaches the ground. The ramp has an angle of 
. The coefficient of kinetic friction for the ramp is 0.1.
What is the work done on the box by friction?
A 2kg box slides down a ramp a distance of two meters before it reaches the ground. The ramp has an angle of
What is the work done on the box by friction?
Tap to reveal answer
Because the force of friction is parallel to the displacement of the box, we can solve for the work done by friction using the following equation.
Now, we need to define the force of friction or 
.
In this formula, 
equals acceleration. In this case, acceleration equals acceleration due to gravity in the x-direction; therefore, use the cosine function.
Insert the value of acceleration due to gravity for 
.
Now, we can solve for the work done by friction.
Because the force of friction is parallel to the displacement of the box, we can solve for the work done by friction using the following equation.
Now, we need to define the force of friction or 
In this formula, 
Insert the value of acceleration due to gravity for 
Now, we can solve for the work done by friction.
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A 
block slides along a smooth surface at a velocity of 
. The block then encounters a rough surface and continues for six meters before coming to a complete stop. What is the force of friction on the block?
A
Tap to reveal answer
This question can be answered using conservation of energy. Having an understanding of the work-energy theorem allows us the knowledge that changing kinetic energy is a form of work, and that work can be a form of energy. In this particular question, the only force acting on the block to slow it down is the force of friction from the rough surface. As we know, work is equal to force times distance.
Because the block started with a certain amount of kinetic energy, and then is brought to a complete stop, all of that kinetic energy is transferred to work done by the force of friction over a distance of six meters.
Because the 
block had a velocity of 
on a smooth surface and ends with a velocity of 
at rest, we can calculate the change in kinetic energy.
The kinetic energy is completely converted to the work done by friction.
We can then use the first equation for work to determine the force of friction.
This question can be answered using conservation of energy. Having an understanding of the work-energy theorem allows us the knowledge that changing kinetic energy is a form of work, and that work can be a form of energy. In this particular question, the only force acting on the block to slow it down is the force of friction from the rough surface. As we know, work is equal to force times distance.
Because the block started with a certain amount of kinetic energy, and then is brought to a complete stop, all of that kinetic energy is transferred to work done by the force of friction over a distance of six meters.
Because the
The kinetic energy is completely converted to the work done by friction.
We can then use the first equation for work to determine the force of friction.
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A student carries a 
stack of books 
along a soccer field. She carries the books at a constant height of one meter above the ground and walks with a constant speed. How much work does the student do on the books?
A student carries a
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Since the height of the books is constant, there is no change in potential energy. Similarly, since the speed of the books is constant, there is no change in kinetic energy. By the work-energy theorem, work is equal to the change in mechanical energy (including potential and kinetic). If there is zero change in energy, the work done must also be zero.
We can also look at this problem in terms of forces. There is zero acceleration on the books, since they do not move in the vertical direction and the horizontal velocity is constant. If there is no acceleration, then there is no force. If there is no force, then there is no work, according to Newton's second law.
Since the height of the books is constant, there is no change in potential energy. Similarly, since the speed of the books is constant, there is no change in kinetic energy. By the work-energy theorem, work is equal to the change in mechanical energy (including potential and kinetic). If there is zero change in energy, the work done must also be zero.
We can also look at this problem in terms of forces. There is zero acceleration on the books, since they do not move in the vertical direction and the horizontal velocity is constant. If there is no acceleration, then there is no force. If there is no force, then there is no work, according to Newton's second law.
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How much work is done to lift a 
block to a point 
above its resting location?
How much work is done to lift a
Tap to reveal answer
The work done is equal to the gravitational potential energy of the block after it has been lifted.
The gravitational potential energy is calculated using the following formula:
We are given the mass and the change in height, and we know the acceleration due to gravity. Using these values, we can solve for the change in potential energy by multiplication.
The work done is equal to the gravitational potential energy of the block after it has been lifted.
The gravitational potential energy is calculated using the following formula:
We are given the mass and the change in height, and we know the acceleration due to gravity. Using these values, we can solve for the change in potential energy by multiplication.
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What is the work done on a 
box that is being pushed with a 
force for 
?
What is the work done on a
Tap to reveal answer
Work is represented by the product of force and displacement:
We are given the force on the box and the distance it travels. Use these values to calculate the work done.
Note that the mass of the box for this problem is irrelevant. There is no vertical displacement, so the force of gravity does not come into play. The only force that matters in this question is the "pushing force."
Work is represented by the product of force and displacement:
We are given the force on the box and the distance it travels. Use these values to calculate the work done.
Note that the mass of the box for this problem is irrelevant. There is no vertical displacement, so the force of gravity does not come into play. The only force that matters in this question is the "pushing force."
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Two students (student X and student Y) lift a boulder vertically from point A to point B. Student X directly lifts the boulder from point A to point B, whereas student Y uses a pulley to lift the boulder. This allows student Y to apply a force (
) that is three times smaller than the force applied by student X (
). Both students apply force upwards and take the same amount of time to complete this task.
The vertical distance between point A and point B is 
.
Student Z uses an inclined plane to lift the boulder and exerts only one third of 
. The work performed by student Z and student X is .
Two students (student X and student Y) lift a boulder vertically from point A to point B. Student X directly lifts the boulder from point A to point B, whereas student Y uses a pulley to lift the boulder. This allows student Y to apply a force (
The vertical distance between point A and point B is
Student Z uses an inclined plane to lift the boulder and exerts only one third of
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Work is generally considered a path function. This means that the amount of work depends on the path taken.
Work in a gravitational field, however, is not a path function; it is a state function. This means that work is independent of the path taken in a gravitational field. This occurs because the force due to gravity acts only in one direction (downwards); therefore, the work performed with or against gravity will only depend on the vertical distance travelled.
In this question both students lift the boulder along the same vertical distance (a distance of 
). Student Z might have travelled a larger total distance by using the inclined plane, but the smaller 
compensates for the larger distance in the work formula.
Work is generally considered a path function. This means that the amount of work depends on the path taken.
Work in a gravitational field, however, is not a path function; it is a state function. This means that work is independent of the path taken in a gravitational field. This occurs because the force due to gravity acts only in one direction (downwards); therefore, the work performed with or against gravity will only depend on the vertical distance travelled.
In this question both students lift the boulder along the same vertical distance (a distance of
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Two children are playing on an icy lake. Child 1 weighs 50kg, and child 2 weighs 38kg. Child 1 has a backpack that weighs 10kg, and child 2 has a backpack that weighs 5kg.
Over the course of the afternoon, they collide many times. Four collisions are described below.
Collision 1:
Child 1 starts from the top of a ramp, and after going down, reaches the lake surface while going and subsequently slides into a stationary child 2. They remain linked together after the collision.
Collision 2:
Child 1 and child 2 are sliding in the same direction. Child 2, moving at , slides into child 1, moving at .
Collision 3:
The two children collide while traveling in opposite directions at each.
Collision 4:
The two children push off from one another’s back, and begin moving in exactly opposite directions. Child 2 moves with a velocity of .
Consider that child 1 adds 5kg of books to her backpack. She then prepares to slide down the ramp to the lake surface. If she slides down the ramp but stops herself after she has lost 2m in vertical height, how much work does the force of gravity perform on her to get her to her new position? Ignore assume the ramp is frictionless.
Two children are playing on an icy lake. Child 1 weighs 50kg, and child 2 weighs 38kg. Child 1 has a backpack that weighs 10kg, and child 2 has a backpack that weighs 5kg.
Over the course of the afternoon, they collide many times. Four collisions are described below.
Collision 1:
Child 1 starts from the top of a ramp, and after going down, reaches the lake surface while going
Collision 2:
Child 1 and child 2 are sliding in the same direction. Child 2, moving at
Collision 3:
The two children collide while traveling in opposite directions at
Collision 4:
The two children push off from one another’s back, and begin moving in exactly opposite directions. Child 2 moves with a velocity of
Consider that child 1 adds 5kg of books to her backpack. She then prepares to slide down the ramp to the lake surface. If she slides down the ramp but stops herself after she has lost 2m in vertical height, how much work does the force of gravity perform on her to get her to her new position? Ignore assume the ramp is frictionless.
Tap to reveal answer
The amount of work performed to get child 1 down the ramp is equal to the amount of kinetic energy gained by her from top to bottom, via the work-energy theorem. Unfortunately, we are not given her kinetic energy. We can't use the 5m/s in the passage, because she stops herself before sliding to the lake surface.
We must instead recognize the potential energy lost is equal to kinetic energy gained. The potential energy that she has lost is given by mgh.
The amount of work performed to get child 1 down the ramp is equal to the amount of kinetic energy gained by her from top to bottom, via the work-energy theorem. Unfortunately, we are not given her kinetic energy. We can't use the 5m/s in the passage, because she stops herself before sliding to the lake surface.
We must instead recognize the potential energy lost is equal to kinetic energy gained. The potential energy that she has lost is given by mgh.
← Didn't Know|Knew It →
Two children are playing on an icy lake. Child 1 weighs 50kg, and child 2 weighs 38kg. Child 1 has a backpack that weighs 10kg, and child 2 has a backpack that weighs 5kg.
Over the course of the afternoon, they collide many times. Four collisions are described below.
Collision 1:
Child 1 starts from the top of a ramp, and after going down, reaches the lake surface while going and subsequently slides into a stationary child 2. They remain linked together after the collision.
Collision 2:
Child 1 and child 2 are sliding in the same direction. Child 2, moving at , slides into child 1, moving at .
Collision 3:
The two children collide while traveling in opposite directions at each.
Collision 4:
The two children push off from one another’s back, and begin moving in exactly opposite directions. Child 2 moves with a velocity of .
Before collision 3 takes place, suppose that child 1 starts at the top of a ramp that is 3m long, and to the lake surface.
With what speed does child 1 hit the lake surface? Ignore friction and air resistance.
Two children are playing on an icy lake. Child 1 weighs 50kg, and child 2 weighs 38kg. Child 1 has a backpack that weighs 10kg, and child 2 has a backpack that weighs 5kg.
Over the course of the afternoon, they collide many times. Four collisions are described below.
Collision 1:
Child 1 starts from the top of a ramp, and after going down, reaches the lake surface while going
Collision 2:
Child 1 and child 2 are sliding in the same direction. Child 2, moving at
Collision 3:
The two children collide while traveling in opposite directions at
Collision 4:
The two children push off from one another’s back, and begin moving in exactly opposite directions. Child 2 moves with a velocity of
Before collision 3 takes place, suppose that child 1 starts at the top of a ramp that is 3m long, and
With what speed does child 1 hit the lake surface? Ignore friction and air resistance.
Tap to reveal answer
This is an application of the work-energy theorem. The amount of work done by the force of gravity to move child 1 down the ramp is equal to force * distance, as well as equal to the amount of kinetic energy picked up down the ramp.
The force down the ramp is equal to mg*sin(30o) = 600J * 0.5 = 300J
This is an application of the work-energy theorem. The amount of work done by the force of gravity to move child 1 down the ramp is equal to force * distance, as well as equal to the amount of kinetic energy picked up down the ramp.
The force down the ramp is equal to mg*sin(30o) = 600J * 0.5 = 300J
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What is the work done by gravity if a 50kg block is pushed along a 40m track?
What is the work done by gravity if a 50kg block is pushed along a 40m track?
Tap to reveal answer
The work done by any force that is perpendicular to the displacement is equal to zero. Since the block is moving horizontally, the net force in the vertical direction will be equal to zero; therefore, work done by either gravity or the normal force will be equal to zero.
The work done by any force that is perpendicular to the displacement is equal to zero. Since the block is moving horizontally, the net force in the vertical direction will be equal to zero; therefore, work done by either gravity or the normal force will be equal to zero.
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An object of mass 100g moves in circular path of radius 0.5m, under the influence of a 15N force directed radially inwards towards the center of the path. How much work is done by this force as the object moves one quarter of the way around the circle?
An object of mass 100g moves in circular path of radius 0.5m, under the influence of a 15N force directed radially inwards towards the center of the path. How much work is done by this force as the object moves one quarter of the way around the circle?
Tap to reveal answer
Work is given by , the dot product of force and displacement. Since the dot product only sees the components of vectors which are parallel to each other, the dot product of two perpendicular vectors is 0. Force is directed radially inward, while displacement is directed tangent to the circumference. At any point along this object's path, the force is perpendicular to displacement, so is simply 0.
Work is given by
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A 2kg box slides down a ramp a distance of two meters before it reaches the ground. The ramp has an angle of 30o. The coefficient of kinetic friction for the ramp is 0.1.
What is the work done by gravity on the box?
A 2kg box slides down a ramp a distance of two meters before it reaches the ground. The ramp has an angle of 30o. The coefficient of kinetic friction for the ramp is 0.1.
What is the work done by gravity on the box?
Tap to reveal answer
Work is determined using the equation . Here, is the force applied, is the displacement of the object, and is the angle of the force relative to the movement of the object. Since gravity is acting on the box, we can solve for the force of gravity causing the movement of the box. Note that in this case refers to the angle between gravity and the box's path; thus, the angle will be 60o, rather than 30o.
Notice how the work done by gravity is equal to the potential energy of the box at the top of the ramp. This is because mechanical energy is conserved in the system; thus, we can set the two equations equal to each other.
Work is determined using the equation
Notice how the work done by gravity is equal to the potential energy of the box at the top of the ramp. This is because mechanical energy is conserved in the system; thus, we can set the two equations equal to each other.
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A 2kg box slides down a ramp a distance of two meters before it reaches the ground. The ramp has an angle of 30o. The coefficient of kinetic friction for the ramp is 0.1.
What is the work done by the normal force on the box?
A 2kg box slides down a ramp a distance of two meters before it reaches the ground. The ramp has an angle of 30o. The coefficient of kinetic friction for the ramp is 0.1.
What is the work done by the normal force on the box?
Tap to reveal answer
Remember that the work done by a force on an object is dependent on the angle of the force to the object's displacement. The normal force acts perpendicularly to the ramp, which means it has an angle of 90o with respect to the box's displacement.
Because , the total work done on the box by the normal force is .
Remember that the work done by a force on an object is dependent on the angle of the force to the object's displacement. The normal force acts perpendicularly to the ramp, which means it has an angle of 90o with respect to the box's displacement.
Because
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A 2kg box slides down a ramp a distance of two meters before it reaches the ground. The ramp has an angle of . The coefficient of kinetic friction for the ramp is 0.1.
What is the work done on the box by friction?
A 2kg box slides down a ramp a distance of two meters before it reaches the ground. The ramp has an angle of
What is the work done on the box by friction?
Tap to reveal answer
Because the force of friction is parallel to the displacement of the box, we can solve for the work done by friction using the following equation.
Now, we need to define the force of friction or .
In this formula, equals acceleration. In this case, acceleration equals acceleration due to gravity in the x-direction; therefore, use the cosine function.
Insert the value of acceleration due to gravity for .
Now, we can solve for the work done by friction.
Because the force of friction is parallel to the displacement of the box, we can solve for the work done by friction using the following equation.
Now, we need to define the force of friction or
In this formula,
Insert the value of acceleration due to gravity for
Now, we can solve for the work done by friction.
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A block slides along a smooth surface at a velocity of . The block then encounters a rough surface and continues for six meters before coming to a complete stop. What is the force of friction on the block?
A
Tap to reveal answer
This question can be answered using conservation of energy. Having an understanding of the work-energy theorem allows us the knowledge that changing kinetic energy is a form of work, and that work can be a form of energy. In this particular question, the only force acting on the block to slow it down is the force of friction from the rough surface. As we know, work is equal to force times distance.
Because the block started with a certain amount of kinetic energy, and then is brought to a complete stop, all of that kinetic energy is transferred to work done by the force of friction over a distance of six meters.
Because the block had a velocity of on a smooth surface and ends with a velocity of at rest, we can calculate the change in kinetic energy.
The kinetic energy is completely converted to the work done by friction.
We can then use the first equation for work to determine the force of friction.
This question can be answered using conservation of energy. Having an understanding of the work-energy theorem allows us the knowledge that changing kinetic energy is a form of work, and that work can be a form of energy. In this particular question, the only force acting on the block to slow it down is the force of friction from the rough surface. As we know, work is equal to force times distance.
Because the block started with a certain amount of kinetic energy, and then is brought to a complete stop, all of that kinetic energy is transferred to work done by the force of friction over a distance of six meters.
Because the
The kinetic energy is completely converted to the work done by friction.
We can then use the first equation for work to determine the force of friction.
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