Question

An $8.0$kg crate is pulled $5.0$m up a $30°$incline by a rope angled $18°$above the incline. The tension in the rope is $120$N, and the crate’s coefficient of kinetic friction on the incline is $0.25$.

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?

Short answer

a). work done by tension is $570.6$J, by gravity is $-196$J and by normal force is $0$J

b). increase in thermal energy is$37.8$J

Step-by-step solution

Step 1. Given information

We have,

An $8.0$kg crate is pulled $5.0$m up a $30°$incline by a rope angled $18°$above the incline. The tension in the rope is $120$N, and the crate’s coefficient of kinetic friction on the incline is $0.25$.

Step 2. Find work done by tension, gravity and normal force

We can use $W=F∆r\cos \alpha$, where $\alpha$is angle between force $F$and direction. In case of tension, that angle is $18°$, in case of gravity is $90°+30°=120°$and $90°$for normal force.

Work done by tension

$$\begin{gathered} W_{tension}=F_{tension}∆r\cos 18° \\ {W}_{tension}=120N·5m·0.95 \\ {W}_{tension}=570.6J \end{gathered}$$

Work done by gravity

$$\begin{gathered} W_{gravity}=F_{gravity}∆r\cos 120° \\ {W}_{gravity}=(8kg·9.8m/s^2)·5m(-0.5) \\ {W}_{gravity}=-196J \end{gathered}$$

Work done by normal force

$$\begin{gathered} W_{normal}=F_{normal}∆r\cos 90° \\ {W}_{normal}=0J \end{gathered}$$

Step 3. Find increase in thermal energy

Thermal energy in this case due to friction.

$$\begin{gathered} m{a}_{perpendiculartocratesmovement}=0=F_{normal}+F_{tension}\sin 18°-F_{gravity}\cos 30° \\ {F}_{normal}=mg\cos 30°-F_{tension}\sin 18° \\ {F}_{normal}=8kg·9.8m/s^2·0.86-120N·0.31 \\ {F}_{normal}=30.24N \end{gathered}$$

Then increase in thermal energy is

$$\begin{gathered} ∆E_{thermal}={\mu }_k{F}_{normal}∆r \\ =0.25·30.24N·5m \\ =37.8J \end{gathered}$$