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A block of mass 30 kg is pulled up a slope (diagram below) with a constant speed by applying a force of 200 N parallel to the slope. A and B are the initial and final positions of the block. Calculate the force of friction offered by the surface AB.

A block of mass 30 kg is pulled up a slope with a constant speed by applying a force of 200 N parallel to the slope. A and B are the initial and final positions of the block. Calculate the force of friction offered by the surface AB. Physics Competency Focused Practice Questions Class 10 Solutions.

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Answer

Given, mass of block (m) = 30 kg

Force applied (F) = 200 N

Work done along AB :
WAB = F x SAB
= 200 x 3 = 600 N m = 600 J

Work done along CB:
WCB = m x g x SCB
= 30 x 10 x 1.5 = 450 J

Let, friction force = f

Work done by friction WF =f x 3

WAB=WCB+WFWABWCB=WFWABWCB=f×3f=WABWCB3f=6004503f=1503f=50 NW{AB} = W{CB} + W{\text F} \\[1em] W{AB} -W{CB} = W{\text F} \\[1em] W{AB} -W{CB} = \text f\times 3 \\[1em] \text f = \dfrac{W{AB} -W{CB}}{3} \\[1em] \text f = \dfrac{600-450}{3} \\[1em] \text f = \dfrac{150}{3} \\[1em] \text f = 50 \text{ N} \\[1em]

Hence, the force of friction offered by the surface AB is 50 N.

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