Work is equal to the change in energy of the system. We are given the weight of the box and the vertical displacement, which will allow us to calculate the change in potential energy. This will be the total work required to move the box against gravity.

The remaining work that the man exerts must have been used to counter the force of friction acting against his motion.

Now we know the work performed by friction. Using this value, we can work to solve for the force of friction and the coefficient of friction. First, we will need to use a second formula for work:

In this case, the distance will be the distance traveled along the surface of the incline. We can solve for this distance using trigonometry.

We know the work done by friction and the distance traveled along the incline, allowing us to solve for the force of friction.

Finally, use the formula for frictional force to solve for the coefficient of friction. Keep in mind that the force on the box due to gravity will be equal to .

Plug in our final values and solve for the coefficient of friction.

In order for mass 1 to remain stationary while mass 2 is experiencing free fall, the magnitude of the force of static friction acting on mass 1 must be greater than the magnitude of the force of tension in the rope that is created by mass 2. The force of tension in the rope is equal to the gravitational force that is created by mass 2. This force can be found using the equation:

The minimum coefficient of static friction would create a force in the opposite direction that has the same magnitude as the force found above. Force of static friction is equal to the coefficient static friction between an object and surface multiplied by the normal force exerted by the object being analyzed. This is represented by the following equation:

Where is the coefficient of static friction and N is normal force. Normal force in this problem would be equal to the mass of object 1 multiplied by the gravity constant. By setting the 2 equations above equal to each other, minimum coefficient of static friction can be found.

After plugging in given values,

Because the block is traveling on an inclined ramp and experiences friction, the two main forces at work here are a component of gravitational force and kinetic frictional force.

The total force acting on the block can be written as , where is the component of gravitational force in the direction of motion, and is the kinetic frictional force.

We know that forces can be expressed as , and kinetic friction can be expressed as . Hence our total force can be expressed as . The normal force can be further expressed as

The component of gravity parallel to the ramp is given by , while the component of gravity perpendicular to the ramp is given by .

Our total force can hence be expressed as .

Thus,

and

Because the mass of object 1 is given, the force acting on it must be found in order to figure out how fast it is accelerating. These 3 variables are related by the equation:

There are 2 forces acting on mass 1. The first force is the tension in the rope. This tension is equal to the magnitude of the gravitational force created by object 2. This can be found using the equation:

where g is the gravity constant which is equal to .

The other force acting on the object is friction. The force of friction can be found by multiplying the normal force exerted by the object on the surface that it is resting on by the coefficient of friction. This is demonstrated by the equation:

Where is the coefficient of friction and N is the normal force. Static friction is the force of friction acting on an object that is not moving. because the magnitude of the force of static friction would not be as great as the magnitude of the force of tension in the rope the object would be moving. Because the object is moving, kinetic friction would be the other force that is acting on mass 1. Kinetic friction represents the friction exerted on an object that is moving on a surface.

Because there are 2 forces acting on this object, a free body diagram can be made to illustrate these forces.

T represents tension in the rope while F represents friction. The total net force can be found by subtracting the force of friction from the magnitude of tension in the rope.

Acceleration of mass 1 can now be figured out because the force acting on the object has been found and the mass is given.

We can use the equation for conservation of energy to solve this problem.

The work in this scenario is done by friction. The only term we can remove from this equaiton is final potential energy. Substituting expressions for each term, we get:

We need to determine initial height and the normal force of the skier before we can solve for the coefficient of friction.

Substitute these into the original equation:

Canceling out mass and rearranging for the coefficient of friction, we get:

Plug in our given values to solve:

We can use one of the kinematic equations to our advantage:

We are given the player's speed and the distance the player slide into home plate. What is implied is that the player's final velocity . We can also rewrite the acceleration in terms of the player's mass and the forces acting on him (which is only friction)

Substitute this into the kinematic equation and solve for the coefficient of kinetic friction,

We can use the expression for conservation of energy to solve this problem:

The problem statement tells us that the sledder is initially at rest, so we can eliminate initial kinetic energy. Also, if we assume that the bottom of the hill has a height of 0, we can eliminate final potential energy to get:

(1)

From here we need to determine what is going to contribute to work. The only extra piece of information we have in this problem is that there is an average frictional force. Therefore that will be the only source of work. Let's began expanding each term, going from left to right:

(2)

Note that we didn't mark the height as an initial height because we assumed that the bottom of the hill has a height of 0.

Where d is the length of the hill. We can calculate this distance using the height and angle of the hill.

Rearranging for distance:

Substituting this back into our expression for work, we get:

(3)

Now our final term:

(4)

Now substituting expression 2, 3, and 4 into expression 1, we get:

The reason we are subtracting friction is because it is removing energy from the system. Another way we could have written our initial expression would be to have work on the final state and then friction would be positive.

Rearranging for the frictional force:

We have all of these values, so time to plug and chug:

In order to determine the normal force, we are going to need the angle of the hill, which we can calculate from the gravitational force that we are given. The vertical force of gravity is:

Its component in the direction of the sledder's movement is:

If you're wondering why sin was used, think about the situation practically. As the angle grows, the hill gets steeper, and there will be more gravitational force in the direction of acceleration. Sine is the function that gets larger as the angle gets larger.

No rearranging for the angle:

Plugging in the values from the problem statement:

Now we can use the FUN equation to determine the kinetic frictional force:

Now we need an expression for the normal force, which is the gravitational force multiplied by cosine:

Once again, if you're unsure why we used cosine, think about the situation practically. As the hill gets flatter (angle decreased), the normal force grows, and cos is the function that gets larger as the angle gets smaller. Now substituting this into the FUN equation, we get:

We have all of our values, so we can solve the problem:

Use the following equations:

The force of friction will point in the opposite direction as the direction of travel, so it will have a negative sign.

Since the energy will be zero when the vehicle has stopped.

Solve for , the stopping distance.

, the mass is not in the solution. Indeed, the extra energy due to mass is cancelled out due to extra friction resulting from the higher mass.

You need to have a firm understanding of static friction in order to answer this question correctly.

The main concept covered in this question is that static fricton between two objects has a max and a min, and can have any value between the max and min.

The max static frictional force can be calculated by the equation:

where mu is the coefficient of friction and N is the normal force

In this problem we get:

However, this is not the answer. The problem statement says that the man is pushing with a force of 35N. Since this is less than the max force calculated, the box will not move. Therefore, the force applied from friction is equal to the force applied by the man, 35N. If these forces were not balanced, then there would be a net force in the opposite direction of the man pushing and the box would accelerate due to friction; this is not possible.