Chapter 10, Problem 69A

To determine

To write: A physics problem based on real-life objects for the given equation.

Explanation of Solution

Given:

The given equation is,

  $\left(12.5\text{N}\right)d=\frac{1}{2}\left(6.0\text{kg}\right){\left(1.10\text{m/s}\right)}^2-\frac{1}{2}\left(6.0\text{kg}\right){\left(0.05\text{m/s}\right)}^2$

Formula used:

According to work energy theorem,

  $\begin{array}{l}W=\frac{1}{2}m{v}_f{}^2-\frac{1}{2}m{v}_i{}^2\\ F\times d=\frac{1}{2}m{v}_f{}^2-\frac{1}{2}m{v}_i{}^2\end{array}$

Where,

F is the force applied.

d is the distance travelled or displacement.

W is work done.

m is the mass of body.

  $v_f$ is the final velocity.

  $v_i$ is the initial velocity.

From the given equation, it is clearly observed thatthe equation is based on work energy theorem.

So,

Force applied on body is $F=12.5\text{N}$

Mass of the body is $m=6.0\text{kg}$

Final velocity of body is, $v_f=1.1\text{0 m/s}$

Initial velocity of the body is, $v_i=0.05\text{ m/s}$

Variable d is unknown in the equation. So, the problem will framed in such a way that distance travelled should be found out using the given equation.

Consider the problem based on real-life objects,

“A 6 kg small ball is rollingon road initially with a speed of 0.05 m/s at point A. Net force of 12.5 N is constantly applied by boy on a ball in the direction of its movement and thus speed reached to 1.10 m/s at point B. What distance covered by ballfrom point A to point B?”