A machine pulls a 46 kg trunk 3.0 m up a 42o ramp at constant velocity, with the machine's force on the trunk directed parallel to the ramp. The coefficient of kinetic friction between the trunk and the ramp is 0.36. What are (a) the work done on the trunk by the machine's force and (b) the increase in thermal energy of the trunk and the ramp?

the answers are

W = 1271.256 J

= 361.81 J

Explanation:

The mass of the trunk (m) = 46kg

angle between the ramp and the horizontal (θ) = 42°

The trunk is pulled at a displacement (d) up the ramp = 3m

The coefficient of kinetic friction between the trunk and the ramp = 0.36

The trunk is moving with a constant velocity therefore the net force on it is zero, therefore the force required to move the trunk must be equal to the summation of forces opposing the trunk

The two forces opposing the trunk are

We have to calculate the machine's force which is equal to sum of the

= θ

μ× N

N = mgcosθ

μθ

F = mg (sinθ + μcosθ)

F = 46× 9.8 (sin42 + 0.36×cos42)

F= 450.8 (0.67 +0.27)

F = 450.8 × 0.94

F = 423.752N

to calculate the workdone on the trunk by the machine force

The workdone on the trunk is  = W = F × dcosΘ

Θ = 0° because the trunk is directed parallel to the ramp

W =423.752× 3 cos 0

W = 423.752 ×3

W = 1271.256 J

(b)  the increase in thermal energy of the trunk and the ramp

Friction converts mechanical energy into thermal energy, so multiplying the frictional force with the distance gives the thermal energy generated by the trunk

× d

= μθ ×d

= 0.36 ×46×9.8×cos42×3

= 361.81 J

i think the answer is tomato juice? not entirely sure though