Question

A man sits with his back against the back of a chair, and he pushes a block of mass $m=2kg$straight forward on a table in front of him, with a constant force F=30 N, moving the block a distance D=0.3 m. The block starts from rest and slides on a low-friction surface. (a) How much work does the man do on the block? (b) What is the final kinetic energy K of the block? (c) What is the final speed V of the block? (d) How much time $\nabla t$does this process take? (e) Consider the system of the man plus the block: how much work does the chair do on the man? (f) What is the internal energy of the man?

Now suppose that the man is sitting on a train that is moving in a straight line with speed $V=15m/s$, and you are standing on the ground as the train goes by, moving to your right. From your perspective (that is, in your reference frame), answer the following questions: (g) What is the initial speed localid="1657950350828" $v_i$of the block? (h) What is the final speed$v_f$of the block? (i) What is the initial kinetic energy of the block? (j) What is the final kinetic energy$k_f$of the block? (k) What is the change in kinetic energy$\nabla K=K_f-K_i$, how does this compare with the change in kinetic energy in the man’s reference frame? (l) How far does the block move$\left(\nabla x\right)$? (m) How much work does the man do on the block, and how does this compare with the work done by the man in his reference frame and with $\nabla K$in your frame? (n) How far does the chair move? (o) Consider the system of the man plus the block: how much work does the chair do on the man, and how does this compare with the work done by the chair in the man’s reference frame? (p) What is the internal energy change of the man, and how does this compare with the internal energy change in his reference frame?

Short answer

(a) Work done by man on block is 9J .

(b) Final kinetic energy of the block is 9 J .

(c)Final speed of the blockis $3m/s$.

(d)Process $0.2s$takestime.

(e)Work done by the chair on the man is0.

(f) Internal energy of the man -9 J .

(g)Initial speed of the blockis $15m/s$.

(h) The final speed of the block is $18m/s$.

(i) Initial kinetic energy of the block is 225 J .

(j)Final kinetic energy of the blockis 324 J .

(k)Change in kinetic energy is 99J .

(l)The block moves by3.3 distance.

(m)Work one by the man on the blockis 99 J .

(n) The chair moves by 3m distance.

(o) Work done by the chair on the man is 0.

(p) Internal energy change of the man is -99 J.

Step-by-step solution

Identification of the given data

The given data can be listed below as follows,

• The mass of the block is, m=2kg
• The force applied on the block is, F=30 N
• The distance moved by the block is, d=0.3m
• The speed of the train is,$V=15m/s$

Explanation of the Work-Energy Principle and Impulse

The change in kinetic energy of a particle is the same as that of the work done on the particle by all the forces. It can be expressed as follows,

$$\begin{gathered} \mathit{W}\mathbf{=}\mathbf{\nabla }\mathit{K} \\ \mathit{W}\mathbf{=}{\mathit{K}}_{\mathbf{f}}\mathbf{-}\mathit{K}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \end{gathered}$$

Here$K_f$is the final kinetic energy of the particle,$K_i$is the initial kinetic energy of the particle, and W is the work done that is equal to the product of the force applied and the distance moved by the particle.

The result of the multiplication of the force applied to an object and the time for which the force is applied is called the impulse. Impulse can also be determined by obtaining the change in momentum. It is expressed as follows,

$$\begin{gathered} I=F\times \nabla t \\ m\nabla v=F\times \nabla t........\left(2\right) \end{gathered}$$

Here, F is the force applied,$\nabla t$is the time consumed, is the mass of the object, and m is the change in the velocity of the object.

Determination of the work done by the man on the block

(a)

Write the expression for the work done by the man on the block by taking the product of force and distance moved by the block,

$W=F\times d$

Substitute all the values from the given data in the above expression,

$$\begin{gathered} W=30N\times 0.3m \\ =9N.m\times \frac{1J}{1N.m} \\ =9J \end{gathered}$$

Thus, the work done by the man on the block is 9J .

Determination of the final kinetic energy of the block

(b)

Write the expression for the change in kinetic energy using equation (1).

$W=K_f-K_j$

Here, $K_j=0$as initially the block was at rest.

Substitute all the values from the given data in the above expression,

$$\begin{gathered} 9J=K_f-0 \\ 9J=K_f \\ {K}_f=9J \end{gathered}$$

Thus, the final kinetic energy of the block is 9J.

Determination of the final speed of the block

(c)

Write the expression for the final kinetic energy of the block,

$K_f=\frac{1}{2}m{v}^2$

Here, v is the final speed of the block.

Substitute all the values in the above expression,

$$\begin{gathered} 9J\times \frac{1kg.m^2/s^2}{1J}=\frac{1}{2}\left(2kg\right)v^2 \\ {v}^2=9m^2/s^2 \\ v=3m/s \end{gathered}$$

Thus, the final speed of the block is$3m/s$ .

Determination of the time consumed in the process

(d)

Substitute all the values in equation (2) and simplify the expression,

$$\begin{gathered} 2\left(kg\right)\times \left(3m/s\right)=30N\times \frac{1kg.m/s^2}{1N}\times \nabla t \\ \nabla t=\frac{6}{30}s \\ \nabla t=0.2s \end{gathered}$$

Thus, the time consumed in the process is 0.2s.

 Step 7: Determination of the work done by the chair on the man

(e)

It can be observed that while applying force on the block, the man does not travel any distance. So, the work done by the chair on the man is zero.

Thus, the work done by the chair on the man is 0.

Determination of the change in internal energy of the man

(f)

It is known thatthe creation and destruction of any kind of energy is not possible, but it can only be transformed from one kind to another. So, the work done by the man on the block is the same as that of the negative value of the internal energy. It can be shown as follows,

$$\begin{gathered} \nabla U=-W \\ =-9J \end{gathered}$$

Thus, the change in the internal energy of the man is -9J .

Determination of the initial speed of the block

(g)

It is known that the block was at rest initially. So, the initial velocity of the block is zero, that is, u = 0 m/s. But the train on which the block is placed is moving with the velocity ofV = 15 m/s.

So, the expression for the initial speed of the block with respect to train is as follows,

$$\begin{gathered} v_j=u+V \\ =0m/s+15m/s \\ =15m/s \end{gathered}$$

Thus, the initial speed of the block is$=15m/s$ .

Determination of the final speed of the block

(h)

It is known that the final velocity of the block was v = 3 m/s when it was not on the train. But the train on which the block is now placed is moving with the velocity ofV = 15 m/s.

So, the expression for the final speed of the block with respect to train is as follows,

$$\begin{gathered} v_f=v+V \\ 3m/s+15m/s \\ =18m/s \end{gathered}$$

Thus, the final speed of the block is$=18m/s$ .

Determination of the initial kinetic energy of the block

(i)

Write the expression for the initial kinetic energy of the block with respect to the train,

$k_i=\frac{2}{1}m{v}_i^2$

Substitute all the values in the above expression,

$$\begin{gathered} k_j=\frac{1}{2}\left(2kg\right)\times {\left(15m/s\right)}^2 \\ =225.\left(kg.m^2/s^2\times \frac{1J}{1kg.m^{2/s^2}}\right) \\ =225J \end{gathered}$$

Thus, the initial kinetic energy of the block is 225 J.

Determination of the final kinetic energy of the block

(j)

Write the expression for the final kinetic energy of the block with respect to the train,

$K_f=\frac{1}{2}m{v}_f^2$

Substitute all the values in the above expression,

$$\begin{gathered} k_f=\frac{1}{2}\left(2kg\right)\times {\left(18m/s\right)}^2 \\ 324.\left(kg.m^2/s^2\times \frac{1J}{1kg.m^2/s^2}\right) \\ =324J \end{gathered}$$

Thus, the final kinetic energy of the block isrole="math" localid="1657953792904" $324J$
.

Determination of the change in kinetic energy of the block

(k)

Substitute all the values in equation (1),

$$\begin{gathered} \nabla K=324J-225J \\ =99.J \end{gathered}$$

Thus, the change in kinetic energy of the block is$99J$ .

The change in kinetic energy of the block is greater than kinetic energy in the man's reference frame.

Determination of the distance moved by the block

(l)

Write the expression for the force that depends on the mass and acceleration of the object,

$F=ma$

Here, m is the mass of the object, and a is the acceleration of the object,

Substitute all the values in the above expression,

$$\begin{gathered} 30N\times \frac{1kg.m/s^2}{1N}=\left(2Kg\right)\times a \\ 30kg.m/s^2=\left(2kg\right)\times a \\ a=15m/s^2 \end{gathered}$$

Write the expression for the second law of motion,

$s=u\left(∆t\right)+\frac{1}{2}a{\left(\nabla t\right)}^2$

Here, s is the distance traveled by the block, u is the initial speed of the block, a is the acceleration, and t is the time taken.

Substitute all the values in the above expression,

$$\begin{gathered} ∆x=15m/s\times \left(0.2s\right)+\frac{1}{2}\left(15m/s^2\right)\times {\left(0.2s\right)}^2 \\ =3.3m \end{gathered}$$

Thus, the distance moved by the block is 3.3 m .

 Step 15: Determination of the work done by the man on the block

(m)

Write the expression for the work done by the man on the block by taking the product of force and distance moved by the block.

$W=F\times ∆x$

Substitute all the values in the above expression.

$$\begin{gathered} w=30N\times 3m \\ =99N.m\times \frac{1J}{1N.m} \\ =99J \end{gathered}$$

Thus, the work done by the man on the block is 99 J.

The work done by the man is greater than the work done in the man's reference frame, and it matches the change in kinetic energy in this frame.

Determination of the distance moved by the chair

(n)

Write the expression for the distance by taking the product of velocity and time taken,

$s=v.∆t$

Here, v is the velocity of the train.

Substitute all the values in the above expression.

$$\begin{gathered} s=15m/s\times \left(0.2s\right) \\ =3m \end{gathered}$$

Thus, the distance moved by the chair is 3m .

Determination of the work done by the chair on the man

(o)

It can be observed that the man and the chair do not travel separately, but they are moving together. So, the work done by the chair on the man is zero.

Thus, the work done by the chair on the man is 0.

Work done by the chair on the man is equal to the work done in the man's reference frame.

Determination of the change in internal energy of the man

(p)

It is known thatthe creation and destruction of any kind of energy is not possible, but it can only be transformed from one kind to another. So, the work done by the man on the block is the same as that of the negative value of the internal energy. It can be shown as follows,

$$\begin{gathered} ∆U=-W \\ =-99J \end{gathered}$$

Thus,the change in the internal energy of the man is -99J.

Change in the internal energy of the manis greater thanthe internal energy of the manin the man's reference frame.