Work, Energy and Power: Tricky Questions Solved (5)

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Tags Work Energy Power

Question

A block of mass m=2m = 2 kg is pulled up a rough incline of angle θ=30∘\theta = 30^\circ by a constant force F=25F = 25 N applied parallel to the incline. The coefficient of kinetic friction is μk=0.2\mu_k = 0.2. The block moves s=5s = 5 m along the incline. Find the net work done on the block and its final speed if it started from rest. Take g=10g = 10 m/s².

Solution — Step by Step

Four forces act on the block: applied force FF (along incline, up), gravity component along incline (mgsin⁡θmg\sin\theta, down), friction (μkmgcos⁡θ\mu_k mg\cos\theta, opposing motion — so down), and normal force (perpendicular, does no work). Only the first three contribute to work along the displacement.

 WF=F s=25×5=125 JW_F = F\,s = 25 \times 5 = 125 \text{ J} Wg=−mgsin⁡θ⋅s=−2×10×0.5×5=−50 JW_g = -mg\sin\theta \cdot s = -2 \times 10 \times 0.5 \times 5 = -50 \text{ J} Wf=−μkmgcos⁡θ⋅s=−0.2×2×10×32×5≈−17.32 JW_f = -\mu_k mg\cos\theta \cdot s = -0.2 \times 2 \times 10 \times \tfrac{\sqrt{3}}{2} \times 5 \approx -17.32 \text{ J} Wnet=125−50−17.32≈57.68 JW_{\text{net}} = 125 - 50 - 17.32 \approx 57.68 \text{ J}

Since the block starts from rest, Wnet=12mv2W_{\text{net}} = \tfrac{1}{2}mv^2.

v=2Wnetm=2×57.682≈7.59 m/sv = \sqrt{\tfrac{2 W_{\text{net}}}{m}} = \sqrt{\tfrac{2 \times 57.68}{2}} \approx 7.59 \text{ m/s}

Net work ≈57.68\approx 57.68 J, final speed ≈7.6\approx 7.6 m/s.

Why This Works

The work-energy theorem says the net work — sum of work by every force, including friction and gravity — equals the change in kinetic energy. We never need acceleration explicitly. This is faster than the Newton’s-second-law route on inclines because we skip the kinematic equation step.

The sign convention is the heart of it: work is positive when force and displacement point the same way, negative when opposite. Friction, on a block moving up, is opposite to motion — so always negative.

Alternative Method

Using Newton’s second law: net force along incline = F−mgsin⁡θ−μkmgcos⁡θ=25−10−3.46=11.54F - mg\sin\theta - \mu_k mg\cos\theta = 25 - 10 - 3.46 = 11.54 N. Acceleration a=11.54/2=5.77a = 11.54/2 = 5.77 m/s². Then v2=0+2×5.77×5⇒v≈7.6v^2 = 0 + 2 \times 5.77 \times 5 \Rightarrow v \approx 7.6 m/s. Same answer, more arithmetic.

Common Mistake

The classic trap is using μkmg\mu_k mg for friction instead of μkmgcos⁡θ\mu_k mg\cos\theta. On a horizontal surface, the normal reaction is mgmg. On an incline, it’s mgcos⁡θmg\cos\theta. Forgetting the cosine here typically loses students 1.5–2 marks in board exams and the entire question in JEE Main.

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