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=(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*5 \\ 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=(1)/(2)mv^2 \\ K=(1)/(2)*6.92*6^2 \\ K=124.56\text{ J} \end{gathered}`

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