Sum

A block of mass 5.0 kg slides down an incline of inclination 30° and length 10 m. Find the work done by the force of gravity.

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#### Solution

\[\text{ Mass of the block, M = 5 kg }\]

\[\text{ Angle of inclination } , \theta = 30^\circ\]

Gravitational force acting on the block,

\[F = mg\]

Work done by the force of gravity depends only on the height of the object, not on the path length covered by the object.

\[\text{ Height of the object , h }= 10 \times \sin30^\circ\]

\[ = 10 \times \frac{1}{2} = 5 m\]

\[ = 10 \times \frac{1}{2} = 5 m\]

\[\therefore \text{ Work done by the force of gravity, w = mgh } \]

\[ = 5 \times 9 . 8 \times 5 = 245 J\]

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