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NEET PHYSICSEasy

Calculate the maximum acceleration of a moving car so that a body lying on the floor of the car remains stationary. The coefficient of static friction between the body and the floor is 0.150.15. (take g=10 m s−2g=10 \text{ m s}^{-2})

A

50 m s−250 \text{ m s}^{-2}

B

1.2 m s−21.2 \text{ m s}^{-2}

C

150 m s−2150 \text{ m s}^{-2}

D

1.5 m s−21.5 \text{ m s}^{-2}

Step-by-Step Solution

  1. Analyze the Physics: For the body to remain stationary relative to the accelerating car, the force responsible for accelerating the body (static friction) must not exceed the limiting friction.
  2. Identify Forces:
  • Downward force: Weight W=mgW = mg.
  • Upward force: Normal reaction N=mgN = mg.
  • Horizontal force required for acceleration aa: F=maF = ma.
  1. Apply Friction Condition: The static friction fsf_s provides the necessary force to accelerate the body along with the car. The condition for the body not to slip is: fs≤fs,max=μsNf_s \le f_{s,max} = \mu_s N ma≤μsmgma \le \mu_s mg a≤μsga \le \mu_s g
  2. Calculate Maximum Acceleration: amax=μsga_{max} = \mu_s g Given μs=0.15\mu_s = 0.15 and g=10 m s−2g = 10 \text{ m s}^{-2}: amax=0.15×10=1.5 m s−2a_{max} = 0.15 \times 10 = 1.5 \text{ m s}^{-2} (Reference: NCERT Class 11, Physics Part I, Chapter 5: Laws of Motion, Section 5.9 and Example 5.7, which solves an identical problem for a box in a train [1]).
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