Question

A body of mass 1.0 kg initially at rest is moved by a horizontal force of 0.5 N on a smooth table. Calculate the work done by the force in 10 s and show that it is equal to the change
in K.E. of the body.

Answer 1

Step 1: Given data

Mass of the body is, $m=1.0kg$

The initial velocity of the body is, $u=0m/s$

Force applied on the body is, $F=0.5N$

Time is, $t=10s$

Step 2: Calculating acceleration

Force is given as,

$F=ma$

where, $m$ is mass and $a$ is acceleration.

Substituting the known values in the above equation of force, we get,

$$\begin{gathered} 0.5N=1.0kg\times a \\ a=\frac{0.5}{1.0} \\ a=0.5m/s^2 \end{gathered}$$

Step 3: Calculating distance through which the body moves

The second equation of motion is given as,

$S=ut+\frac{1}{2}a{t}^2$

where, $S$ is distance, $u$ is initial velocity, $t$ is time and $a$ is acceleration.

Substituting the known values in the above equation, we get,

$$\begin{gathered} S=\left(0m/s\times 10s\right)+\left(\frac{1}{2}\times 0.5m/s^2\times {\left(10s\right)}^2\right) \\ S=0+25 \\ S=25m \end{gathered}$$

Step 4: Calculating work done

Work done is given as,

$W=FS$

where, $F$ is force and $S$ is distance.

Substituting the known values in the above equation, we get,

$$\begin{gathered} W=0.5N\times 25m \\ W=12.5J \end{gathered}$$
Step 5: Calculating the change in kinetic energy of the body

The third equation of motion is given as,

$v^2=u^2+2aS$

where, $v$ is final velocity, $u$ is initial velocity, $a$ is acceleration and $S$ is distance.

Substituting the known values in the above equation, we get,

$$\begin{gathered} v^2={\left(0m/s\right)}^2+2\times 0.5m/s^2\times 25m \\ {v}^2=25 \\ v=5m/s \end{gathered}$$

Kinetic energy is given as,

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

Substituting the known values in the above equation, we get,

$$\begin{gathered} KE=\frac{1}{2}\times 1.0kg\times {\left(5m/s\right)}^2 \\ KE=\frac{1}{2}\times 25 \\ KE=12.5J \end{gathered}$$

Therefore, the work done is $12.5J$ which is equal to the change in kinetic energy of the body. $\left(KE=12.5J\right)$