The work done in dragging a block of mass $5 k g$ on an inclined plane of height $2 m$ is $150$ Joule. The work done against the frictional force will be

200

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150

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100

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50

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Solution

The correct option is C $50$ J
From FBD,
$F = m g sin θ + f$
or $F = m g \frac{h}{x} + f$ as $(sin θ = \frac{h}{x})$
or $F x = m g h + f x$
or $W_{F} = m g h + W_{f}$ as work done $=$ force $\times$ distance.
$150 = 5 \times 10 \times 2 + W_{f}$
or $W_{f} = 150 - 100 = 50 J$

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The work done in dragging a block of mass 5 kg on an inclined plane of height 2 m is 150 Joule. The work done against the frictional force will be

The correct option is C 50 JFrom FBD, F=mgsinθ+for F=mghx+f as(sinθ=hx)or Fx=mgh+fxor WF=mgh+Wf as work done = force × distance. 150=5×10×2+Wfor Wf=150−100=50J

The work done in dragging a block of mass $5 k g$ on an inclined plane of height $2 m$ is $150$ Joule. The work done against the frictional force will be

The correct option is C $50$ J
From FBD,
$F = m g sin θ + f$
or $F = m g \frac{h}{x} + f$ as $(sin θ = \frac{h}{x})$
or $F x = m g h + f x$
or $W_{F} = m g h + W_{f}$ as work done $=$ force $\times$ distance.
$150 = 5 \times 10 \times 2 + W_{f}$
or $W_{f} = 150 - 100 = 50 J$