A block weighing 20 N travels down a smooth | vedclass

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(B) According to the work-energy theorem,the total work done on the block is equal to the change in its kinetic energy.$W_{	ext{gravity}} + W_{	ext{

Vedclass · Accepted Answer

$2.5$

Vedclass · Answer

(B) According to the work-energy theorem,the total work done on the block is equal to the change in its kinetic energy.$W_{	ext{gravity}} + W_{	ext{friction}} = KE_f - KE_i$Since the block starts from rest and comes to rest after moving a distance $x$ on the rough surface,$KE_i = 0$ and $KE_f = 0$.The work done by gravity is $W_{	ext{gravity}} = mgh = (mg)h = 20 \ N 	imes 1.0 \ m = 20 \ J$.The work done by friction is $W_{	ext{friction}} = -f_k \cdot x = -(\mu \cdot mg) \cdot x = -(0.40 	imes 20 \ N) \cdot x = -8x \ J$.Applying the work-energy theorem:$20 - 8x = 0$$8x = 20$$x = \frac{20}{8} = 2.5 \ m$.Thus,the distance the block will move on the rough surface is $2.5 \ m$.

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