A crate of fruit with mass 35.0 kg and specific heat 3650 J/kg · K slides down a ramp inclined at 36.9° below the horizontal. The ramp is 8.00 m long. (a) If the crate was at rest at the top of the incline and has a speed of 2.50 m/s at the bottom, how much work was done on the crate by friction? (b) If an amount of heat equal to the magnitude of the work done by friction goes into the crate of fruit and the fruit reaches a uniform tinal temperature, what is its temperature change?
Solution 33E Problem (a) Step 1: Mass of the crate with fruit m = 35 kg Angle of inclination of the ramp = 36.9° Length of the ramp L = 8 m Speed of the crate at the bottom v = 2.5m/s Specific heat c = 3650 J/Kg.K Step 2: A work is done by the friction to hold the crate on the top. Here the energy is conserved when the crate is on the top. The potential energy(PE) at a height is equal to the sum of kinetic energy and the work done by the friction. 1 2 mgh = mv 2 w f w f = mgh - mv2 2 If the crate started moving the energy stored (PE) is divided into kinetic energy and w f Step 3: To find the height of the ramp Height of the ramp h = sin 36.9° * length of the ramp h = 0.6*8 h = 4.8 m Step 4: 1 2 w f = mgh - m2 w f = 35*9.8*4.8 - 1 *35*2.5 2 2 w = 1537.025 J or 1.54 KJ f The amount of work done by the friction is w f = 1.54 KJ Problem (b) To find the change in temperature T Step 1: Amount of work done by friction = the quantity of heat w f= Q = 1540 J c = 3650 J/Kg.K