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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 ?

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 ?

Grade:12

9 Answers

SHAIK AASIF AHAMED
askIITians Faculty 74 Points
9 years ago
Hello student,
Fs=mgh
F=mgh/s
by substituting values we get F=20N
Thanks and Regards
Shaik Aasif
askIITians faculty
Gaurav
11 Points
6 years ago
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.
Mudit Kesharwani
21 Points
6 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
6 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
6 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.
Om Anil Gonade
9 Points
6 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
Its the actual answer.
Here s=0.2 because force is applied for 0.2 m. 
Prashant
14 Points
6 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
5 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/s2 . now put these values in F=ma. f=0.2X100=20N......
is there any mistake????
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the answer to your question.
 
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

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