Solveeit Logo
Login

Question

Physics Question on work, energy and power

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

A

40 ms−140 \, ms^{ - 1}

B

20 ms−120 \, ms^{ - 1}

C

10 ms−110 \, ms^{ - 1}

D

1030ms−110 \sqrt{30} ms^{ - 1}

Answer

40 ms−140 \, ms^{ - 1}

Explanation

Solution

According to conservation of energy mgH = 12mv2+mgh2\frac{ 1}{ 2} mv^2 + mgh_2 or mg(H−h2)=12mv2mg ( H - h_2 ) = \frac{ 1}{ 2} mv^2 or v=2g(100−20)v = \sqrt{ 2g ( 100 - 20 ) } or v=2×10×80=40 ms−1 v = \sqrt{ 2 \times 10 \times 80 } = 40 \, ms^{ - 1}