Question

A rope is used to pull a 3.57 kgblock at constant speed 4.06 malong a horizontal floor. The force on the block from the rope is 7.68 Nand directed$\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{°}$above the horizontal. What are: (a) the work done by the rope’s force, (b) the increase in thermal energy of the block-floor system, and (c) the coefficient of kinetic friction between the block and floor?

Short answer

1. The work done by the rope’s force is W = 30.1 J.
2. The increase in thermal energy of block-floor system is$∆E_{th}=30.1J$.
3. The coefficient of kinetic friction between block-floor system is ${\mu }_k=0.225$.

Step-by-step solution

Step 1: Given Data

The mass of block, m = 3.57 kg.

The displacement of block along a floor, d = 4.06 m.

The magnitude of force from the rope, F = 7.68 N.

The force of rope is directed at angle $15°$above horizontal.

Determining the concept

Use the equation of work relating force and displacement. The increase in thermal energy will be equal to the work done by applied force as the velocity is constant. To calculate the coefficient of kinetic friction, apply Newton’s second law of motion to the block-earth system.

Formulae are as follow:

$\begin{array}{l}W=∆E_{mec}+∆E_{th}\\ ∆E_{mec}=∆K+∆U\\ ∆E_{th}=fd\\ W=fd\cos \theta \end{array}$

where,$∆K$ is kinetic energy,$∆U$ is potential energy,$∆E_{mec}$ is mechanical energy,$∆E_{th}$is thermal energy, d is displacement, F, fare forces and W is work done.

Step 3(a): Determining the work done by the rope’s force

The equation for work done by applied force is,

$W=fd\cos \theta$

Hence,

$\begin{array}{l}W=7.68\times 4.06\times \cos \left(15\right)\\ W=30.1J\end{array}$

Hence, the work done by the rope’s force is W = 30.1 J.

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

Now,

$∆E_{th}=fd$

When the acceleration of the block is zero,

$fd=W$

So,

role="math" localid="1661400999788" $∆E_{th}=W=30.1J$

Hence, the increase in thermal energy of block-floor system is $∆E_{th}=30.1J$.

Step 5(c): Determining the coefficient of kinetic friction between block-floor system

Draw a free-body diagram for the block-floor system,

Now, write the force equations from this diagram as,

$F\cos \theta -f=ma·······\left(1\right)$

And,

$N+F\sin \theta -mg=ma........\left(2\right)$

Solving equation (1), with a= 0,

$\begin{array}{l}f=F\cos \theta =7.68\times \cos \left(15\right)\\ f=7.42N\end{array}$

And, solving equation (2) for normal force N with a=0,

$\begin{array}{l}N=mg-F\sin \theta \\ =3.57\times 9.81-7.68\times \sin \left(15\right)\\ N=33N\end{array}$

Now, the equation for friction force as,

$f={\mu }_{k}N$

So, the coefficient of kinetic friction is,

$$\begin{gathered} {\mu }_k=\frac{f}{N} \\ =\frac{7.42}{33} \\ {\mu }_k=0.225 \end{gathered}$$

Hence,the coefficient of kinetic friction between block-floor system is${\mu }_k=0.225$.

Therefore, the work done and the increase in thermal energy can be found using the equation of work relating applied force and displacement. The coefficient of kinetic friction is the ratio of friction force and the normal force. Hence, it can be found by solving the force equations.