Question

A horizontal force of magnitude 35.0 Npushes a block of mass 4.00 kgacross a floor where the coefficient of kinetic friction is 0.600. (a) How much work is done by that applied force on the block-floor system when the block slides through a displacement of 3.00 macross the floor? (b) During that displacement, the thermal energy of the block increases by40.0 J. What is the increase in thermal energy of the floor? (c) What is the increase in the kinetic energy of the block?

Short answer

1. The work done by the applied force on the block-floor system W=105 J.
2. The amount of increase in thermal energy of the floor $∆E_{th}\left(floor\right)=30.6J$.
3. The increase in the kinetic energy of the block$∆K=34.4J$

Step-by-step solution

Step 1: Given Data

The mass of block is m = 4.00 kg

The magnitude of horizontal force is F = 35.0 N

The coefficient of kinetic friction is${\mu }_k=0.600$

The displacement of the block across the floor is d = 3.00 m

The increase in thermal energy of the block is$∆E_{th}T\left(block\right)=40.0J$

Determining the concept

To solve this problem, use the equation of work done by the external force to calculate work. For the increase in thermal energy, use the equation of work due to friction force. For the increase in kinetic energy used, the equation for work relating with change in mechanical energy and change in thermal energy.

Formulae are as follow:

$$\begin{gathered} W=F.d \\ ∆E_{th}=f_{k}d \\ W=∆E_{mec}+∆E_{th} \end{gathered}$$

Where,$∆E_{mec}$ is change in mechanical energy, $∆E_{th}$is change in thermal energy, d is displacement, F,$f_k$ are forces and W is work done.

Step 3(a): Determining the work done by the applied force on the block-floor system

Now,

W = F.d

So,

$\begin{array}{l}W=35.0\times 3.00\\ W=105J\end{array}$

Hence, the work done by the applied force on the block-floor system W = 105 J.

Step 4(b): Determining the amount of increase in thermal energy of the floor

Now,

$∆E_{th}=f_{k}d$

Hence, the amount of increase in thermal energy oftheblock-floor system is,

$$\begin{gathered} ∆E_{th}={\mu }_{k}mgd \\ =0.600\times 4.00\times 9.81\times 3.00 \\ ∆E_{th}=70.6J \end{gathered}$$

Now,

$∆E_{th}\left(floor\right)=∆E_{th}-∆E_{th}\left(block\right)$

So,

$$\begin{gathered} ∆E_{th}\left(floor\right)=∆E_{th}-∆E_{th}\left(block\right) \\ =70.6-40.0 \\ ∆E_{th}\left(floor\right)=30.6J \end{gathered}$$

Hence, the amount of increase in thermal energy of the floor, $∆E_{th}\left(floor\right)=30.6J$.

Step 5(c): Determining the increase in the kinetic energy of the block

The work done on the system is equal to the sum of change in the mechanical energy and change in thermal energy of system. So,

$\begin{array}{l}W=∆E_{mec}+A{E}_{th}\\ ∆E_{mec}=W-∆E_{th}=105-70.6\\ ∆E_{mec}=34.4J\end{array}$

As the motion of the block is across the floor, there is no change in the potential energy of the block. So,

$∆E_{mec}=34.4J$

Hence,the increase in the kinetic energy of the block,$∆K=34.4J$

Therefore, the change in kinetic energy of the block and the change in thermal energy of the floor along with the work done can be found using the corresponding equations.