A block of 20 kg mass is pulled up a slope at constant velocity by applying force acting parallel to the slope. If the slope makes an angle of 30 degree with the horizontal, calculate
i) The work done in pulling the load up a distance of 3.0 meters along the slope.
ii) The increase in its potential energy of the block?
Details and Assumptions
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We have to consider the friction case to get the exact answer. We know, tan(theta)= "mu". So, we get (mu)=tan 30 = (1/sqrt3). Then, The total force required to neutralize the downward weight component & friction force becomes= mg(sin 30)+"mu"mg(cos 30) = (20)(9.8)=196 N. So, the net work done by this force is W= (196)(3) J=588 J. Well, Then, the height the block has achieved is (3)(sin 30) = 1.5. So, the increase in potential energy is = mg(1.5)=(20)(9.8)(1.5)=294 J. At last, We sum up these. The answer becomes = 588+294 =882 J