FIGURE 6 shows a 3.0 kg box sliding down with constant speed u = 1.2 ms-1 along a rough plank inclined at an angle 25o with the horizontal. Calculate the rate of work done by the frictional force on the block. (Given g = 9.81 ms-2)
Afr=−mv22+mglsinα=−860 J.A_{fr}=-\frac{mv^2}2+mgl\sin\alpha=-860~J.Afr=−2mv2+mglsinα=−860 J.
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