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Work Energy Theorem

A 1 kg ball i...

Question

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

5.5 J

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6.5 J

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7.5 J

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8.5 J

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Solution

The correct option is C $7.5 J$
Taking ground as reference,
Total mechanical energy of the ball at point $A$ $= U_{A} + K E_{A} = m g h_{A} + 0$
$= (1) \times (10) \times 2 + 0$
$= 20 J$
And, total mechanical energy of ball when it reaches point $C$
$= U_{C} + K E_{C} = m g h_{C} + K E_{C}$
$= K E_{C} + (1 \times 10 \times 1.25)$
$= K E_{C} + 12.5$

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

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A 1 kg ball is released and slides down from point A to point C, as shown in the figure below. If the acceleration due to gravity=10 m/s2, the kinetic energy of the ball when it arrives at point C.

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