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JEE Main 2024 (30 Jan, Shift 2) · Work by a Constant Force

Work by a Constant Force question from JEE Main 2024 (30 Jan, Shift 2)

Q10JEE Main 2024 · 30 JAN · SHIFT 2Medium

Work by a Constant Force

A block of mass 1 kg1\text{ kg} is pushed up a surface inclined to horizontal at an angle of 60∘60^\circ by a force of 10 N10\text{ N} parallel to the inclined surface as shown in figure. When the block is pushed up by 10 m10\text{ m} along inclined surface, the work done against frictional force is:

[g=10 m/s2]\left[g = 10\text{ m/s}^2\right]

Figure

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Answer: (b)

  1. Normal reaction N=mgcos⁡60∘=1×10×12=5N = mg\cos 60^\circ = 1 \times 10 \times \frac{1}{2} = 5 N
  2. Friction f=μN=0.1×5=0.5f = \mu N = 0.1 \times 5 = 0.5 N
  3. Work against friction =fs=0.5×10=5= f s = 0.5 \times 10 = 5 J
  4. So the answer is (b) 5 J.
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