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AP EAMCET · PHYSICS · Laws of Motion

A conveyor belt is moving horizontally with a velocity of \(2 \mathrm{~m} \mathrm{~s}^{-1}\). If a body of mass 10 kg is kept on it, then the distance travelled by the body before coming to rest is
(The coefficient of kinetic friction between the belt and the body is 0.2 and acceleration due to gravity is \(10 \mathrm{~m} \mathrm{~s}^{-2}\) )

  1. A 4 m
  2. B 0 m
  3. C 1 m
  4. D 2 m
Verified Solution

Answer & Solution

Correct Answer

(C) 1 m

Step-by-step Solution

Detailed explanation

\(F_k = \mu_k mg = 0.2 \times 10 \mathrm{~kg} \times 10 \mathrm{~m} \mathrm{~s}^{-2} = 20 \mathrm{~N}\) \(a = \frac{F_k}{m} = \frac{20 \mathrm{~N}}{10 \mathrm{~kg}} = 2 \mathrm{~m} \mathrm{~s}^{-2}\)…
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