A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m which applying the force and the ball goes upto 2 m height further, find the magnitude of the force. Consider g = 10 m/s2 ?

You are raising the ball`s potential energy. During the 0.2 m , you raise only PE, but after that you raise both PE and KE.THEREFOREFs=mghF(0.2)=(0.2)(10)(0.2+2)F=22NThat`s your answer.

Dear students,Fs = mgh Here h will be the sum total of distance of hand and the original distance as here both KE and PE will work,Here h=2+0.2=2.2So, F = mgh/sF=0.2×10×2.2/0.2F=10×2.2F=22N Thanks .....Hope you clearly understand this...For any other query ask me in the comment box

But the answer is 16n and not 20n or 22n please correct the answer and also please provide me with the details of the solution.

 

Fs = mgxHere x is the sum total of distance of hand and the original distance .Here both Kinetic energy and Potential energy will work,So, x=2So, F = mgx/sF=0.2×10×2/0.2F=10×2F=20NHere s=0.2 because force is applied for 0.2 m.