The work done in pulling up a block of wood weighing 2kN for a length of 10 m on a smooth plane inclined at an angle of $15^{\circ}$ with the horizontal is $(s i n 15^{\circ} = 0.259)$

4.36 k J

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5.17 k J

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8.91 k J

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9.82 k J

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Solution

The correct option is B
5.17 k J

Work done = Change in potential energy
$Work done = m g \times h$
$= 2 \times 10^{3} \times l s i n θ$
$= 2 \times 10^{3} \times 10 \times s i n 15^{\circ} = 5176 J = 5.17 k J$

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The work done in pulling up a block of wood weighing 2kN for a length of 10 m on a smooth plane inclined at an angle of 15∘with the horizontal is (sin15∘=0.259)

The correct option is B 5.17 k J Work done = Change in potential energy Work done=mg×h =2×103×l sinθ =2×103×10×sin15∘=5176 J=5.17 kJ

The work done in pulling up a block of wood weighing 2kN for a length of 10 m on a smooth plane inclined at an angle of $15^{\circ}$ with the horizontal is $(s i n 15^{\circ} = 0.259)$

The correct option is B
5.17 k J

Work done = Change in potential energy
$Work done = m g \times h$
$= 2 \times 10^{3} \times l s i n θ$
$= 2 \times 10^{3} \times 10 \times s i n 15^{\circ} = 5176 J = 5.17 k J$