Question:

A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying die force and the ball goes upto 2 m height further, find the magnitude of the force. Consider g = 10 m/$s^{2}$:

Updated On: Jul 5, 2022
  • 4 N
  • 16 N
  • 20N
  • 22N
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The Correct Option is D

Solution and Explanation

The situation is shown in figure. At initial time, the ball is at P, then under the action of a force (exerted by hand) from P to A and then from A to B let acceleration of ball during PA is a m/$s^{2}$[assumed to be constant] in upward A direction and velocity of ball at A is v m/s. Then for PA, $v^{2}=0^{2}+2a\times0.2$ $For AB, \, \quad0=v^{2}-2\times g\times2$ $\Rightarrow\, \quad v^{2}=2g\times2$ From above'equations, $a=10g=100 m /s^{2} $ Then for PA, FBD of ball is F - mg = ma[ F is the force exerted by hand on ball] $\Rightarrow\,\quad$ F = m(g + a) = 0.2(1 lg) = 22N Alternate solution : Using work-energy theorem $W_{mg}+w_{F}=0$ $\Rightarrow-mg\times2.2+F\times0.2=0 \Rightarrow F=22N$
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Concepts Used:

Work-Energy Theorem

The work and kinetic energy principle (also known as the work-energy theorem) asserts that the work done by all forces acting on a particle equals the change in the particle's kinetic energy. By defining the work of the torque and rotational kinetic energy, this definition can be extended to rigid bodies.

The change in kinetic energy KE is equal to the work W done by the net force on a particle is given by,

W = ΔKE = ½ mv2f − ½ mv2i

Where, 

vi → Speeds of the particle before the application of force

vf → Speeds of the particle after the application of force

m → Particle’s mass

Note: Energy and Momentum are related by, E = p2 / 2m.