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Laws of Motion question

2023 · Q159
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Laws of Motion question

2023 · Q159

NEETPhysicsLaws of MotionMCQ+4 / −1

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

(g=10 m s−2)\left(g=10 \mathrm{~m} \mathrm{~s}^{-2}\right)(g=10 m s−2).

  1. A
    150 m s−2150 \mathrm{~m} \mathrm{~s}^{-2}150 m s−2
  2. B
    1.5 m s−21.5 \mathrm{~m} \mathrm{~s}^{-2}1.5 m s−2
  3. C
    50 m s−250 \mathrm{~m} \mathrm{~s}^{-2}50 m s−2
  4. D
    1.2 m s−21.2 \mathrm{~m} \mathrm{~s}^{-2}1.2 m s−2
View written solutionFree

Correct answer: B

To find the maximum acceleration ($a_{\max}$) of the car that allows a body to stay stationary relative to the car, we use the concept of static friction. Static friction ($F_s$) is what keeps the body from sliding on the car's floor. It acts in the opposite direction of the potential movement of the body.

The maximum static friction force is given by $$F_{s_{\max}} = \mu_s \times N$$ where $\mu_s$ is the coefficient of static friction and $N$ is the normal force. In this scenario, the normal force is equal to the gravitational force on the body ($mg$), where $m$ is the mass of the body and $g$ is the acceleration due to gravity.

Since the maximum force of static friction equals the product of the mass and the maximum acceleration ($ma_{\max}$), we have:

$\mu_s mg = ma_{\max}$

By canceling out the mass $m$ on both sides, we get:

$a_{\max} = \mu_s g$

Substituting the given values ($\mu_s = 0.15$ and $g = 10 \,\mathrm{m/s}^2$):

$$a_{\max} = 0.15 \times 10 = 1.5 \,\mathrm{m/s}^2$$

Therefore, the maximum acceleration of the car to ensure the body remains stationary with respect to the car's floor is $1.5 \,\mathrm{m/s}^2$.

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