A block of mass 2.0 kg is pushed down an inclined place of inclination $37^{\circ}$ with a force of 20 N acting parallel to the incline. It is found that the block moves on the incline with an acceleration of 10 $m / s^{2}$ . If the block started from rest, find the work done (a) by the applied force in the first second, (b) by the weight of the block in the first second and (c) by the frictional force acting on the block in the first second. Take g= $10 m / s^{2}$ .

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Solution

$m = 2 k g, θ = 37^{\circ}, F = 20 N, a = 10 m / s e c^{2} (a) t = 1 s e c$

So, $s = u t + \frac{1}{2} a t^{2} = 5 m$

Work done by the applied force

$W = F S c o s θ^{2} = 20 \times 5 = 100 J$

$(b) B C (h) = 5 s i n 37^{\circ} = 3 m$

So, frictional force,

$f = m g s i n θ$

Work done by the friction forces,

$W = f s c o s 0^{\circ} = (m g s i n θ) s$

$= 20 \times 0.60 \times 5 = - 60 J$

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m=2kg,θ=37∘,F=20N,a=10m/sec2(a)t=1sec So, s=ut+12at2=5m Work done by the applied force W = FS cosθ2=20×5=100J (b) BC (h) = 5sin 37∘=3 m So, frictional force, f = mg sinθ Work done by the friction forces, W=fs cos 0∘=(mg sin θ)s =20×0.60×5=−60J