Question

A ball of mass 0.2kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2m while applying the force and the ball goes up to 2m height further, find the magnitude of the force. Consider g=$10m{s^{ - 2}}$
A.20N
B.22N
C.4N
D.16N

Answer 1

Hint: We need to understand the scenario of the question given that when the ball is thrown vertically we must apply some force. Then by using the formula maximum height and the third equation of motion we can calculate the magnitude of the applied force on the ball.

Complete step by step solution:
From the given data:
Mass of the ball =0.2kg
Distance moved by hand=0.2m
Maximum height= 2m
g =$10m{s^{ - 2}}$
When the ball started moving then its velocity is taken as u and when it reaches to a maximum height is taken as ${H_{\max }}$
By using the maximum height formula, we get
$\
  {H_{\max }} = \dfrac{{{u^2}}}{{2g}} \\
  u = \sqrt {2g{H_{\max }}} \\
\implies u = \sqrt {2 \times 10 \times 2} \\
\implies u = 2\sqrt {10} m{s^{ - 1}} \\
\ $
This velocity is supplied to the ball by the hand and initially the hand was at rest, it acquires this velocity in distance of 0.2m
By using the third equation of motion,
${u^2} = 2as$
$\
  a = \dfrac{{{u^2}}}{{2s}} \\
\implies a = \dfrac{{{{\left( {2\sqrt {10} } \right)}^2}}}{{2 \times 0.2}} \\
\implies a = 100m{s^{ - 2}} \\
\ $
Here we need to calculate the magnitude of the force such that the upward force is applied on the ball and hence acceleration and the gravity both are in the same direction
Therefore, by using the newton’s law of motion we get
$\
  F = m\left( {g + a} \right) \\
\implies F = 0.2(10 + 100) \\
\implies F = 22N \\
\ $

So, the correct answer is “Option B”.

Note:
When the body which is at rest the initial velocity is taken as zero.
When the body is moving downward direction the acceleration is taken as positive and when it is moving upward then it is negative.
When the body is at the maximum height then the final velocity is taken as zero.