Question

A $1\text{kg}$ ball is released and slides down from point $\text{A}$ to point $\text{C}$, as shown in the figure below. If the acceleration due to gravity$=10\text{m}/s^2$, the kinetic energy of the ball when it arrives at point $\text{C}$.

• A. $5.5\text{J}$
• B. $6.5\text{J}$
• C. $7.5\text{J}$
• D. $8.5\text{J}$

Answer 1

The correct option is C $7.5\text{J}$

Taking ground as reference,
Total mechanical energy of the ball at point $\text{A}$ $=U_A+K{E}_A=mg{h}_A+0$
$=\left(1\right)\times \left(10\right)\times 2+0$
$=20\text{J}$
And, total mechanical energy of ball when it reaches point $\text{C}$
$=U_C+K{E}_C=mg{h}_C+K{E}_C$
$=K{E}_C+(1\times 10\times 1.25)$
$=K{E}_C+12.5$

From the principal of mechanical energy conservation, total energy of ball at point A equals the total energy of the ball at point C.
$⟹(U_A+K{E}_A)=(U_C+K{E}_C)$
$⟹20=K{E}_C+12.5$
$⟹K{E}_C=(20-12.5)=7.5\text{J}$
Hence , the kinetic energy of the ball when it reach at point $\text{C}$ is $7.5\text{J}$