A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of $60^{{}^{\circ}}$ by a force of 10 N parallel to the inclined surface as shown in figure. When the block is pushed up by 10 m along inclined surface, the work done against frictional force is : $\left[g=10 \mathrm{~m} / \mathrm{s}^2\right]$

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A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of

60^{{}^{\circ}}

by a force of 10 N parallel to the inclined surface as shown in figure. When the block is pushed up by 10 m along inclined surface, the work done against frictional force is :

\left[g=10 \mathrm{~m} / \mathrm{s}^2\right]

A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of 60^ ^ by a force of 10 N parallel to

Quantrex Academy · Accepted Answer

Correct answer: B. 5 J

Mass of block, m=1 kg Force of parallel inclined surface, F=10 N Work done against frictional force =μ_2 N × 10=μ Mg 60 × 10=0.1 × 5 × 10=5 J

A block of mass 1 kg is pushed up a surface inclined to horizontal...

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