Question

A person holds a ball of mass 0.25 kg in his hand and throws it, so that it leaves his hand with a speed of 12 m s$^{-1}$. In the process, if his hand moved through a distance of 0.9 m, then the net force acted on the ball is
Identify the correct option from the following:

• A. 40 N
• B. 20 N
• C. 30 N
• D. 10 N

Hint:

Use the work-energy theorem or kinematic equations to find the net force when a body is accelerated over a distance.

Answer 1

The Correct Option is B

Step 1: Calculate the work done on the ball
Mass $m = 0.25$ kg, final speed $v = 12$ m/s, initial speed $u = 0$, distance $d = 0.9$ m. Kinetic energy gained by the ball = $\frac{1}{2} m v^2 = \frac{1}{2} \times 0.25 \times 12^2 = 0.125 \times 144 = 18$ J. Work done $W = F \cdot d$, so $F \cdot 0.9 = 18$, $F = \frac{18}{0.9} = 20$ N. Step 2: Alternative approach using kinematics
Use $v^2 = u^2 + 2as$: $12^2 = 0 + 2 \times a \times 0.9$, $144 = 1.8a$, $a = \frac{144}{1.8} = 80$ m/s$^2$. Net force $F = ma = 0.25 \times 80 = 20$ N. Step 3: Match with options
The net force 20 N matches option (2).