# 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 ?

SHAIK AASIF AHAMED
7 years ago
Hello student,
Fs=mgh
F=mgh/s
by substituting values we get F=20N
Thanks and Regards
Shaik Aasif
Gaurav
11 Points
5 years ago
You are raising the balls 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=22NThats your answer.
Mudit Kesharwani
21 Points
4 years ago
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=2.2
So, F = mgh/s
F=0.2×10×2.2/0.2
F=10×2.2
F=22N
Naman
22 Points
4 years ago
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
Dyth
11 Points
4 years ago
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.
9 Points
4 years ago

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
So, F = mgh/s
F=0.2×10×2/0.2
F=10×2
F=20N
Here s=0.2 because force is applied for 0.2 m.
Prashant
14 Points
4 years ago
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.
Ramu Pentakota
26 Points
3 years ago
friends and teachers  ,,, you all gave different answers ,, so i am getting confused.
to avoid confusion ,i done this question using kinematics …
the ball raised 2m so …the velocity of the ball at the time of leaving the hand sould be $\sqrt{}2gh=2\sqrt{10}$.
the hand is raised by 0.2 m . now by applying $v^{2}-u^{2}=2as$ where s=0.2m, and u=0 and v=$2\sqrt{10}$.
therefore a=100m/s. now put these values in F=ma. f=0.2X100=20N......
is there any mistake????
one year ago
Dear student,

The work done by the hand will be stored as the Potential energy of the ball at its topmost point
The maximum height attained by ball from the initial position = 2 + 0.2 = 2.2m
Hence,
F*s = mgh
or, F = mgh/s
= 0.2 * 10 * 2.2 / 0.2 = 22N
Hope it helps.
Thanks and regards,
Kushagra