Question

What will be it’s final speed and kinetic energy after it has been moving for t=5sec

Answer 1

Given data:

* The force applied on the block is,

$F_a=12.5\text{ N}$

* The frictional force acting on the block is,

$F_r=4.2\text{ N}$

* The acceleration of the block is,

$a=1.2ms^{-2}$

Solution:

The net force acting on the block is,

$\begin{gathered} F_{\text{net}}=F_a-F_r \\ F_{\text{net}}=12.5-4.2 \\ F_{\text{net}}=8.3\text{ N} \end{gathered}$

According to Newton's second law, the net force in terms of the mass and acceleration of the block is,

$\begin{gathered} F_{\text{net}}=ma \\ m=\frac{F_{\text{net}}}{a}_{} \end{gathered}$

Substituting the known values,

$\begin{gathered} m=\frac{8.3}{1.2} \\ m=6.92\text{ kg} \end{gathered}$

As the block is initially at rest, thus, the initial speed of the block is u = 0 m/s.

By the kinematics equation, the final speed of the block after t = 5 seconds is,

$\begin{gathered} v-u=at \\ v-0=1.2\times5 \\ v=6\text{ m/s} \end{gathered}$

Thus, the final speed of the block after t = 5 seconds is 6 meters per second.

The kinetic energy of the block after t = 5 seconds is,

$\begin{gathered} K=\frac{1}{2}mv^2 \\ K=\frac{1}{2}\times6.92\times6^2 \\ K=124.56\text{ J} \end{gathered}$

Thus, the kinetic energy of the block after t = 5 seconds is 124.56 joule.