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 $2.5 \ m$.

Column-$I$	Column-$II$
$(1)$ Potential Energy $=$ Kinetic Energy	$(a)$ At height $\frac{2h}{3}$
$(2)$ Potential Energy $= 2 \times$ Kinetic Energy	$(b)$ Constant at every point
	$(c)$ At height $\frac{h}{2}$

Similar Questions

The velocity-time graph for a body of mass $10 \ kg$ is shown in the figure. The work done on the body in the first two seconds of motion is:

$A$ body of mass $2 \, kg$ is released from point $A$. Its velocity at point $B$ is $4 \, m/s$,and it comes to rest at point $C$. The work done against friction is ............. $J$.

$A$ body falls freely from a height $h$ in a gravitational field. Match the following:

Column-$I$ Column-$II$

$(1)$ Potential Energy $=$ Kinetic Energy $(a)$ At height $\frac{2h}{3}$

$(2)$ Potential Energy $= 2 \times$ Kinetic Energy $(b)$ Constant at every point

$(c)$ At height $\frac{h}{2}$

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