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Q. A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It rolls down a smooth surface to the ground, then climbs up another hill of height 30 m and finally rolls down to a horizontal base at a height of 20 m above the ground. The velocity attained by the ball is

Work, Energy and Power

Solution:

According to conservation of energy
mgH = $ \frac{ 1}{ 2} mv^2 + mgh_2 $
or $ mg ( H - h_2 ) = \frac{ 1}{ 2} mv^2 $
or $ v = \sqrt{ 2g ( 100 - 20 ) } $
or $ v = \sqrt{ 2 \times 10 \times 80 } = 40 \, ms^{ - 1}$

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