Explanation
Given:
The mass of the bucket m=20 kg
The height of the building h=20 m
Work done by the worker=Potential energy of bucket at the top of the building
∴W=mgh
⇒W=20×9.8×20
⇒W=3920J
Thus, the required work done is 3.92kJ
K.E1 = K.E2
12 M1 v21 = 12 M2 v22
Given that,
Spring constant k=100
Mass m=10kg
Stretching x=2m
We know that,
The work done is
W=12kx2
W=12×100×4
W=200J
Hence, the work done is 200 J
Two identical balls are released from positions as shown. They collide elastically on horizontal surface. Ratio of heights attained by A & B after collision is( All surface are smooth,neglect energy loss at M & N)
Work done in stretching wire
=12YAL2L=12F.l
Where
L= length of wire
l= increase in length
We know that
E=12×F×AL
=12×200×10−3
=0.1J
Hence (A) option is correct.
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