A crate of fruit with a mass of 30.5kg and a specific heat capacity of 3800J / (kg? K) slides 7.70m down a ramp inclined at an angle of 36.2degrees below the horizontal.

Part A

If the crate was at rest at the top of the incline and has a speed of 2.35m/s at the bottom, how much work Wf was done on the crate by friction?

Use 9.81m/s^2 for the acceleration due to gravity and express your answer in joules.

Wf =

-1280J

Answer: - 1,277 J

Explanation:

When no non-conservative forces are present, the total mechanical energy (sum of the kinetic energy and the potential energy) must be conserved.

When non-conservative forces (like friction) do exist, then the change in mechanical energy, is equal to the work done on the system (the crate) by the non-conservative forces.

So, we can write the following expression:

∆K + ∆U = WFNC

If the crate starts from rest, this means that the change in kinetic energy, is simply the kinetic energy at the bottom of the ramp:

∆K = ½ m v2 = ½. 30.5 kg. (2.35) 2 m2 = 84.2 J (1)

Regarding gravitational potential energy, if we take the bottom of the ramp as the zero reference level, we have:

∆U = 0 - m. g. h = - m. g. h

In order to get the value of the height of the ramp h, we can apply the definition of the sinus of an angle:

sin θ = h/d, where d is the distance along the ramp = 7.7 m.

Replacing the values, and solving for h, we have:

h = 7.7 sin 36.2º = 4.55 m

So, replacing the value of h in the equation for ∆U:

∆U = 0 - (30.5 kg. 9.81 m/s2. 4.55 m) = - 1,361 J (2)

Adding (1) and (2):

∆K + ∆U = 84.2J - 1,361 J = - 1,277 J

As we have already said, this value is equal to the work done by the non-conservative forces (friction in this case).