A body of mass 2 kg is at rest at the bottom | vedclass

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(C) Given: Mass $m = 2 \ kg$,length of incline $S = 8 \ m$,height $h = 1 \ m$,coefficient of friction $\mu = 0.2$,$g = 10 \ m/s^2$.From the geometry,$

Vedclass · Accepted Answer

$51$

Vedclass · Answer

(C) Given: Mass $m = 2 \ kg$,length of incline $S = 8 \ m$,height $h = 1 \ m$,coefficient of friction $\mu = 0.2$,$g = 10 \ m/s^2$.From the geometry,$\sin 	heta = h/S = 1/8$.Using $\sin^2 	heta + \cos^2 	heta = 1$,we get $\cos 	heta = \sqrt{1 - (1/8)^2} = \sqrt{63/64} = \sqrt{63}/8 \approx 7.937/8 \approx 0.992$.To move the body at a constant speed,the applied force $F$ must balance the gravitational component and the frictional force: $F = mg \sin 	heta + \mu mg \cos 	heta$.The work done $W = F 	imes S = (mg \sin 	heta + \mu mg \cos 	heta) 	imes S$.$W = mgS \sin 	heta + \mu mgS \cos 	heta = mgh + \mu mgS \cos 	heta$.$W = (2 	imes 10 	imes 1) + (0.2 	imes 2 	imes 10 	imes 8 	imes 0.992) = 20 + 31.74 = 51.74 \ J$.Rounding to the nearest integer provided in the options,the correct answer is $51 \ J$.

$A$ body of mass $2 \ kg$ is at rest at the bottom of an inclined plane of length $8 \ m$ and height $1 \ m$ as shown in the figure. If the coefficient of friction is $0.2$,what is the work done in moving the body from the bottom to the top of the incline (in $J$)? (Take $g = 10 \ m/s^2$)

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