Question

A worker pushed a 27 kgblock9.2 malong a level floor at constant speed with a force directed$\mathbf{32}\mathbf{°}$below the horizontal. If the coefficient of kinetic friction between block and floor was 0.20, what were (a) the work done by the worker’s force and (b) the increase in thermal energy of the block– floor system?

Short answer

1. The work done by the force of worker is $W=5.6\times {10}^2\mathrm{J}$
2. The increase in thermal energy of the block–floor system is $∆E_{th}=5.6\times {10}^2\mathrm{J}$

Step-by-step solution

Step 1: Given Data

The mass of block is,$m=27\mathrm{kg}$.

The displacement of a block is $d=9.2\mathrm{m}$.

The applied force is directed$32°$below the horizontal.

The coefficient of kinetic friction is, $\mu =0.20$.

Determining the concept

Use the equation of the work done related with force and displacement. Calculate the force applying Newton’s second law of motion. The thermal energy gets created due to the friction force, so the increase in thermal energy is equal to the work done against the friction force.

Formulae are as follow:

$$\begin{gathered} W=Fd\cos \theta \\ {F}_{net}=ma \end{gathered}$$

where, m is mass, a is an acceleration, d is displacement, F is force and W is work done.

Step 3(a): Determining the work done by the force of the worker

Draw a free body diagram for the block,

From this figure, applying Newton’s second law of motion to the horizontal direction,

$$\begin{gathered} F_{net}=ma \\ F\cos \left(32\right)-f=ma \end{gathered}$$

But,

$a=0$,

So,

$$\begin{gathered} F\cos \left(32\right)-{\mu }_{k}N=0 \\ F\cos \left(32\right)={\mu }_k\left[mg+F\sin \left(32\right)\right] \end{gathered}$$

Solving this equation for applied force F,

$$\begin{gathered} F=\frac{{\mu }_{k}mg}{\cos \left(32\right)-{\mu }_k\sin \left(32\right)}\left(\mathrm{where}\mathrm{\mu }=0.2,\mathrm{m}=27\mathrm{kg}\right) \\ F=71.39\mathrm{N} \end{gathered}$$

The equation for work done is,

$W=Fd\cos \theta$

So,

$$\begin{gathered} W=71.39\times 9.2\times \cos \left(32\right) \\ W=557\mathrm{J}\approx 5.6\times {10}^2\mathrm{J} \end{gathered}$$

Hence, the work done by the force of worker is $W=5.6\times {10}^2\mathrm{J}$

Step 4(b): Determining the increase in thermal energy of the block–floor system

The change in thermal energy is given by,

$$\begin{gathered} ∆E_{th}=fd \\ ∆E_{th}={\mu }_k\left[mg+F\sin \left(32\right)\right]\times d \\ ∆E_{th}=0.20\left[27\times 9.81+71.39\sin \left(32\right)\right]\times 9.2 \\ ∆E_{th}=557\mathrm{J}\approx 5.6\times {10}^2\mathrm{J} \end{gathered}$$

Hence, the increase in thermal energy of the block–floor system is $∆E_{th}=5.6\times {10}^2\mathrm{J}$

Therefore, the work done by the force of the worker and the increase in the thermal energy of the block-floor system can be found using the equation of work.