MCAT Physical - Work | Practice Hub

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?

Explanation

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.

$W=F\cdot d$

Now, we need to define the force of friction or $F_{f}$ .

$F_f=\mu\cdot m \cdot a$

In this formula, equals acceleration. In this case, acceleration equals acceleration due to gravity in the x-direction; therefore, use the cosine function.

$F_{f}=(0.1)\cdot(2kg)\cdot (g\cdot \cos(30^o))$

Insert the value of acceleration due to gravity for .

$F_{f}=0.1\cdot 2kg \cdot -9.8\tfrac{m}{s^2}\cdot \cos(30^o)$

Now, we can solve for the work done by friction.

$W_{friction} =F_f\cdot d$

$W_{friction}=(-1.697N)\cdot(2m)$

$W_{friction} = -3.39J$

What is the work done by gravity if a 50kg block is pushed along a 40m track?

Explanation

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.

$W_g=(10\frac{m}{s^2})(0m)=0J$

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 $5\frac{m}{s}$ 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 $10\frac{m}{s}$ , slides into child 1, moving at $2\frac{m}{s}$ .

Collision 3:

The two children collide while traveling in opposite directions at $10\frac{m}{s}$ 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 $+8\frac{m}{s}$ .

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.

$5.4\frac{m}{s}$

$0.9\frac{m}{s}$

$55\frac{m}{s}$

$900\frac{m}{s}$

$9.8\frac{m}{s}$

Explanation

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

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?

Explanation

Work is determined using the equation $W = Fdcos\Theta$ . Here, is the force applied, is the displacement of the object, and $cos\Theta$ 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 $\Theta$ refers to the angle between gravity and the box's path; thus, the angle will be 60o, rather than 30o.

$W = Fdcos\Theta = (2kg)(10\frac{m}{s^{2}})(2 meters)cos(60^o)$

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.

$\Delta U_{g} = Fdcos\Theta$

A block slides along a smooth surface at a velocity of $2\frac{m}{s}$ . 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?

$\frac{1}{3}\textup{N}$

$\frac{1}6\textup{N}{}$

$3\textup{N}$

$6\textup{N}$

Explanation

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.

$W=\Delta KE=\frac{1}{2}m(\Delta v)^2$

Because the block had a velocity of $2\frac{m}{s}$ on a smooth surface and ends with a velocity of $0\frac{m}{s}$ at rest, we can calculate the change in kinetic energy.

$\Delta KE=\frac{1}{2}(1.0kg)(0\frac{m}{s}-2\frac{m}{s})^2=\frac{1}{2}(1.0kg)(4\frac{m^2}{s^2})$

$\Delta KE=2J$

The kinetic energy is completely converted to the work done by friction.

$W=\Delta KE=2J$

We can then use the first equation for work to determine the force of friction.

$F_f=\frac{2J}{6m}=\frac{1}{3}N$

How much work is done to lift a $\small 2kg$ block to a point $\small 1m$ above its resting location?

Explanation

The work done is equal to the gravitational potential energy of the block after it has been lifted.

$W=\Delta PE$

The gravitational potential energy is calculated using the following formula:

$\Delta PE = mg\Delta h$

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.

$\Delta PE=(2kg)(9.8\frac{m}{s^2})(1m-0m)$

$\Delta PE=19.6J$

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?

47J

4.8J

11.8J

7.5J

Explanation

Work is given by $\small \small \small W = \overrightarrow{F} \cdot \overrightarrow{d}$ , 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 $\small \small \overrightarrow{F}\cdot\overrightarrow{d}$ is simply 0.

What is the work done on a $\small 45kg$ box that is being pushed with a $\small 100N$ force for $\small 5m$ ?

Explanation

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."

A student carries a $\small 10kg$ stack of books $\small 50m$ 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?

Explanation

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.

$W=\Delta PE+\Delta KE$

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.

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:

Collision 2:

Child 1 and child 2 are sliding in the same direction. Child 2, moving at $10\frac{m}{s}$ , slides into child 1, moving at $2\frac{m}{s}$ .

Collision 3:

The two children collide while traveling in opposite directions at $10\frac{m}{s}$ 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 $+8\frac{m}{s}$ .

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.

Explanation

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.

Work

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