Question

A collie drags its bed box across a floor by applying a horizontal force of 8.0 N. The kinetic frictional force acting on the box has magnitude 5.0 N . As the box is dragged through 0.70 malong the way, what are (a) the work done by the collie’s applied force and (b) the increase in thermal energy of the bed and floor?

Short answer

1. The work done by the force of the coolie is $W=5.6\mathrm{J}$.
2. The increase in thermal energy of the block–floor system is $∆E_{th}=3.5\mathrm{J}$

Step-by-step solution

Step 1: Given Data

The displacement of a block is, d = 0.70 m

The applied drag force of collie is, F = 8.0 N

The kinetic friction force is, f =5.0 N

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 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 applied force of coolie

The equation of work done is,

$$\begin{gathered} w=Fd\cos \theta \\ W=8.0\times 0.70\cos \left(0\right) \\ W=5.6\mathrm{J} \end{gathered}$$

Hence, the work done by the force of the coolie is, $W=5.6\mathrm{J}$.

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

The change in thermal energy is given by,

$$\begin{gathered} ∆E_{th}=fd \\ ∆E_{th}=5.0\times 0.70 \\ ∆E_{th}=3.5\mathrm{J} \end{gathered}$$

Hence,the increase in thermal energy of the block–floor system is,$∆E_{th}=3.5\mathrm{J}$

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