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Question

# The following data was obtained for a body of mass 1 kg dropped from a height of 5 metres : Distance above ground Velocity 5 m 3.2 m 0 m 0 m/s 6 m/s 10 m/s Show by calculations that the above data verifies the law of conservation of energy (Neglect air resistance). (g = 10 m/s2).

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Solution

## Mass of the body, (m1) = 1 kg Acceleration due to gravity, (g) = 10 m/s2 We can calculate the potential energy as, Potential energy = (Weight of body) × (Vertical distance) Now, we can find the kinetic energy as, $K.E=\frac{1}{2}m{v}^{2}$ We have three cases: Case – 1 Height (h) = 5 m Velocity (v) = 0 m/s So, Potential energy = (Weight of body) × (Vertical distance) 1 × 10 × 5 = 50 J And, $K.E=\frac{1}{2}m{v}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}×\left(1\right)×\left(0\right)\mathrm{J}\phantom{\rule{0ex}{0ex}}=0\mathrm{J}$ So total energy at this instant, P.E + K.E = (50 + 0) J = 50 J Case – 2 Height (h) = 3.2 m Velocity (v) = 6 m/s So, Potential energy = (Weight of body) × (Vertical distance) = (1) × (10) × (3.2) J = 32 J And, $K.E=\frac{1}{2}m{v}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}×\left(1\right)×\left(6{\right)}^{2}\mathrm{J}\phantom{\rule{0ex}{0ex}}=18\mathrm{J}$ So total energy at this instant, P.E + K.E = (32 + 18) J = 50 J Case – 3 Height (h) = 0 m Velocity (v) = 10 m/s So, Potential energy = (Weight of body) × (Vertical distance) = (1) × (10) × (0) J = 0 J And, $K.E=\frac{1}{2}m{v}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(1\right)\left(10{\right)}^{2}\mathrm{J}\phantom{\rule{0ex}{0ex}}=50\mathrm{J}$ So total energy at this instant, = P.E + K.E = (0 + 50) J = 50 J We can observe that total energy in all the above cases is same, which satisfies the law of conservation of energy.

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