Sankalp NEET Full Test-2 Question-40 Solution

Question 40: 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 0.15\left(\mathrm{~g}=10 \mathrm{~m} \mathrm{~s}^{-2}\right).

(1) 1.5 \mathrm{~m} \mathrm{~s}^{-2}

(2) 50 \mathrm{~m} \mathrm{~s}^{-2}

(3) 1.2 \mathrm{~m} \mathrm{~s}^{-2}

(4) 150 \mathrm{~m} \mathrm{~s}^{-2}

Answer: Option (1)

Explanation:

For the body to remain stationary with respect to the car, the frictional force must provide the necessary acceleration.

The maximum static friction force is given by f_{\max}=\mu_s N.

Here, the normal reaction is N=mg.

Thus, the maximum frictional force is f_{\max}=\mu_s mg.

This frictional force provides the maximum possible acceleration a_{\max} to the body, so

f_{\max}=ma_{\max}.

Therefore, ma_{\max}=\mu_s mg.

Canceling m from both sides, we get

a_{\max}=\mu_s g.

Substituting the given values,

a_{\max}=0.15 \times 10.

Hence, a_{\max}=1.5 \mathrm{~m\,s^{-2}}.

Therefore, the correct answer is Option (1).

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