JEE MAIN - Physics (2006 - No. 45)
A particle of mass $$100g$$ is thrown vertically upwards with a speed of $$5$$ $$m/s$$. The work done by the force of gravity during the time the particle goes up is
$$-0.5J$$
$$-1.25J$$
$$1.25J$$
$$0.5J$$
Explanation
Kinetic energy at point of throwing is converted into potential energy of the particle during rise.
$$K.E = {1 \over 2}m{v^2} = {1 \over 2} \times 0.1 \times 25 = 1.25\,J$$
$$W = - mgh = - \left( {{1 \over 2}m{v^2}} \right) = - 1.25\,J$$
$$\left[ \, \right.$$ As we know, $$mgh = {1 \over 2}m{v^2}$$ by energy conservation $$\left. \, \right]$$
$$K.E = {1 \over 2}m{v^2} = {1 \over 2} \times 0.1 \times 25 = 1.25\,J$$
$$W = - mgh = - \left( {{1 \over 2}m{v^2}} \right) = - 1.25\,J$$
$$\left[ \, \right.$$ As we know, $$mgh = {1 \over 2}m{v^2}$$ by energy conservation $$\left. \, \right]$$
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