A 35 kg trunk is pushed 7.2 m at constant speed up a 30° incline by a force along the plane. The coefficient of kinetic friction between the trunk and the incline is 0.25.

Question

1) Calculate the work done by the weight of the truck

2)Calculate the work done by the applied force

3) How much energy was dissipated by the frictional force acting on the trunk

Answer 1

Hi Al C.!

Typically, drawing a free body diagram is the best place to start with problems involving forces. We want to visualize which forces we have acting on the system and how they are acting. Draw your diagram with the box traveling along the plane of the ramp, the force due to gravity pointing directly downward, the applied force pointing along the plane, the force due to kinetic friction downward along the plane and the normal force perpendicular to the surface.

Now, it is probably helpful if we tilt the axis by 30 degrees so that it lines up perfectly with the normal force on the y-axis, and the applied and frictional forces on the x-axis.

If were to draw the axis in the "normal" convention, we would align only the force due to gravity along the y-axis and no other forces along any axes. By tilting it equal to the angle of the ramp, we have aligned three forces with the axes. We have "eliminated" splitting these forces into components. We need only to split the force due to gravity into its component vectors. Another look at our new alignment.

Now, we are ready to setup our Newton's 2nd Law equations.

∑F_y = n - F_g = ma = 0

n - F_g = ma = 0 => n = F_g

The force due to gravity is the only force that need split into components. The component in the y-direction takes the form:

Fg_y = mgcosθ

We can then make the following statement:

n - F_g = ma = 0 => n = F_g = mgcosθ

We will keep this equation for now and use it later. Let's now consider our horizontal direction:

∑F_y = F_app - f_k - F_g = ma = 0

Because the question informs us that the box moves with constant speed, we can infer that the acceleration is equal to 0. Let's now rearrange our equation to solve for the applied force:

F_app - f_k - F_g = 0 => F_app = f_k + F_g

=> f_k = µ_kn = µ_k(mgcosθ)

=> Fg = mgsinθ

F_app = µ_k(mgcosθ) + mgsinθ = mg(µ_kcosθ + sinθ)

<=> (35kg)(9.8m*s^-2)[(0.25)(cos30 + sin30)]

F_app = 245.8N

Now that we have ascertained the value of the applied force, we can plug this into our equation for work.

W_app = F_appΔx

<=> (245.8N)(7.2m)

W_app = 1769.5J

Next, calculating the work done by the force due to gravity. In this case, we will apply our work equation:

W = F_gΔycosθ = F_ghcosθ

We are not given the value for the distance the box moves in the y-direction. However, we can find it easily and plug it in:

sinθ = height/distance => height = distance*sinθ

W_g = F_gΔycosψ = F_g(distance*sinθ)cosψ = mg(distance*sinθ)cosψ

Let's make sure we know where these Greek letters are coming from!

1. θ is the angle between the force due to gravity and the normal
2. ψ is the angle between the force due to gravity and the displacement

<=> (35kg)(9.8m*s^-1)(7.2m*sin30)cos(180)

W_g = -1234.8J

Note that the answer is negative because the work is done against gravity.

Finally, the work done by friction is also a straightforward application of the work equation:

W_f = f_kΔxcosΨ = µ_k(mgcosθ)ΔxcosΨ

Let's again be careful about our angles here, and make sure we know where all of these Greek letters are coming from!

1. θ is the angle between the force due to gravity and the normal
2. Ψ is the angle between the displacement and the force due to kinetic friction

W_f = µ_k(mgcosθ)ΔxcosΨ

<=> (0.25)(35kg)(9.8m*s^-2)(cos30)(7.2m)(cos180)

W_f = -534.7J

Notice that the difference between the work done by the applied force and that of weight is equal to the work done by friction. This is so because the change in the kinetic energy is zero. Therefore, the summation of work must be equal to zero.

Hope this help!

Cheers