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Question

A man drops a 10 kg rock from the top of a 5 m ladder. What is its speed just before it hits the ground? What is its kinetic energy when it reaches the ground? (g = 10 m/s2)

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Solution

Mass of the body dropped, (m) = 10 kg Height from which the body is dropped, (h) = 5 m Initial velocity of the ball, (u) = 0 m/s Acceleration due to gravity, (g) = 10 m/s2 Let the final velocity of the ball when it just reaches the ground be (v) We will use the third equation of motion to find the velocity of the body when it just reaches the ground, v2 = u2 + 2as So, $v=\sqrt{0+2\left(10\right)\left(5\right)\mathrm{m}/\mathrm{s}}\phantom{\rule{0ex}{0ex}}=\overline{)10\mathrm{m}/\mathrm{s}}$ Now, we can find the kinetic energy as, $K.E=\frac{1}{2}m{v}^{2}$ Therefore, $K.E=\frac{1}{2}\left(10\right)\left(10{\right)}^{2}\mathrm{J}\phantom{\rule{0ex}{0ex}}=\overline{)500\mathrm{J}}$ So, the kinetic energy of the moving object is 500 J.

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