How Does Friction Affect Energy Conversion on an Inclined Plane?

[rvhockey]

Two 10 kilogram boxes are connected by a massless string that passes over a massless frictionless pulley as shown above. The boxes remain at rest, with the one on the right hanging vertically and the one on the left 2.0 meters from the bottom of an inclined plane that makes an angle of 60° with the horizontal. The coefficients of kinetic friction and static friction between the Ieft hand box and the plane are 0.15 and 0.30, respectively. You may use g = 10 m/s², sin 60° = 0.87, and cos 60° = 0.50.
a. what is the tension T in the string?
b. Draw and label all forces acting on the box that is on the plane.
c. Determine the magnitude of the frictional force acting on the box on the plane (when it is at rest)

The string is then cut and the box slides down the plane.
d. Determine the amount of mechanical energy that is converted into thermal energy during the slide to the bottom.
e. Determine the kinetic energy of the box when it reaches the bottom of the plane.


KE= mv²/2
NonConservative Work = deltaKE + delta PE
Work = Force * distance




a. I said 100N, but i am not completely sure. I only said it because i thought the tension was constant throughout the whole string, and since T = mg in the hanging block because it was at rest, i said T = 100

b. i have normal force = mgcos60
and mgsin60 pointing down the slope with frictional force

c. I said 13N because it is at rest and mgsin60+force of friction should equal T, but using the coeffictients of friction and the normal force I'm getting 15N ?

d. I said it should be 26J because of 13N times 2m, but i remembered that the frictional force would be kinetic, so F_N*coefficient of kinetic = 7.5N, so the thermal energy would be 15J ?

e. I said 148J using -26J as the W_NC but i got 159J using the other frictional force as the nonconservative force.?/?

I know how to get the answers I'm just confused as to which ones are right. Any help or insight on this would be great. Thanks.

 

[Doc Al]

a. I said 100N, but i am not completely sure. I only said it because i thought the tension was constant throughout the whole string, and since T = mg in the hanging block because it was at rest, i said T = 100

Good thinking!

b. i have normal force = mgcos60
and mgsin60 pointing down the slope with frictional force

What about the force from the string?

c. I said 13N because it is at rest and mgsin60+force of friction should equal T,

Good!

but using the coeffictients of friction and the normal force I'm getting 15N ?

Careful here. μN gives the maximum possible static friction force between the two surfaces--but to find the actual friction force requires more information (as above).

d. I said it should be 26J because of 13N times 2m,

Oops. The 13N was a static friction force--not relevant once the box moves.

but i remembered that the frictional force would be kinetic, so F_N*coefficient of kinetic = 7.5N, so the thermal energy would be 15J ?

Much better.

e. I said 148J using -26J as the W_NC but i got 159J using the other frictional force as the nonconservative force.?/?

Using static friction was an error, but you have it now.

 

[thomate1]

I cannot provide an answer key or confirm the accuracy of specific answers. However, I can offer some guidance on how to approach and solve these types of problems.

First, it is important to carefully read the question and identify all the given information and what is being asked. In this case, we are given the masses, distances, angles, and coefficients of friction, and we are asked to find the tension in the string, the forces acting on the box on the plane, the magnitude of the frictional force, the amount of energy converted into thermal energy, and the kinetic energy of the box.

Next, we can use the given information and apply the principles of Newton's laws of motion and conservation of energy to solve the problem. For part a, we can use the equation T = mg to find the tension in the string, which in this case would be 100N.

For part b, we can draw a free-body diagram and label all the forces acting on the box on the plane. These include the normal force, the weight of the box, the frictional force, and the component of the weight parallel to the plane. We can then use the equations for force and motion to find the magnitude and direction of these forces.

For part c, we can use the coefficient of static friction and the normal force to find the maximum frictional force, which in this case would be 15N. Remember that the frictional force is always equal and opposite to the applied force, so in this case, it would be equal to the component of the weight parallel to the plane.

For part d, we can use the work-energy theorem to find the amount of energy converted into thermal energy during the slide. This involves calculating the work done by the frictional force, which is equal to the frictional force multiplied by the distance the box slides. In this case, it would be 15N times 2m, which is 30J.

For part e, we can use the conservation of energy principle to find the kinetic energy of the box when it reaches the bottom of the plane. This involves finding the initial potential energy of the box at the top of the plane and subtracting the work done by non-conservative forces (frictional force) to find the final kinetic energy. In this case, the initial potential energy is mgh, where h is the height of the inclined plane, and the work done by non-conservative forces is