Question

A 7.8 kg crate is pulled 5.4 m up a 30^o incline by a rope angled 19^o above the incline. The tension in the rope is 120 N and the crate's coefficient of kinetic friction on the incline is 0.24.

a) How much work is done by tension, by gravity, and by the normal force?

b) What is the increase in thermal energy of the crate and incline?

Answer 1

Sol::

Given ::

m= 7.8 kg
x = 5.4 m
Ѳ = 30°
Ѳ= 19°
Tension (T) =120 N
Coefficient of kinetic friction= 0.24.

A)

Work done by Tension
Wt = T*Δr
= T*Δx*Cos19°
= 120*5.4*(cos 19)
= 612.696 J
≈ 613 J

Work done by Gravity
Wg = F*Δr
= m*g*Δx*Cos(90+30)
= - 7.8*9.81*5.4*(sin 30)
= -206.598 J
≈ -207 J
  
Work done by Normal force
Wn = F(normal)*zero
= 0 J
B)

The increase in thermal energy of the crate and incline
= Work done against the force of friction
= [0.24*{7.8*9.81*(cos 30) - 120*(sin 19)}]*5.4
= 35.249 J
≈ 35.25 J

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