Question:

A vehicle moving at $36\, km/hr$ is to be stopped by applying brakes in the next $5\, m$. If the vehicle weighs $2000\, kg,$ determine the average force that must be applied on it

Updated On: Jun 7, 2024
  • $10^4 N$
  • $ 2 \times 10^4 N$
  • $ 3 \times 10^4 N$
  • $5 \times 10^3 N$
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The Correct Option is B

Solution and Explanation

Given, initial velocity of vehicle $u=36\, km / h$ $=10\, m / s$ Stopping distance of vehicle $s=5\, m$ and mass of vehicle $m=2000\, kg$ So, from equation of motion $v^{2}-u^{2}=2 a s$ since, final velocity of vehicle $v=0$ and acceleration of vehicle $=-a$ (acceleration) $\Rightarrow -u^{2} =-2 a s $ $\Rightarrow a =\frac{u^{2}}{2 s} $ $=\frac{10 \times 10}{2 \times 5}=10\, m / s ^{2}$ The average force for braking $F=m a=2000 \times 10=2 \times 10^{4} N$
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Concepts Used:

Newtons Laws of Motion

Newton’s First Law of Motion:

Newton’s 1st law states that a body at rest or uniform motion will continue to be at rest or uniform motion until and unless a net external force acts on it.

Newton’s Second Law of Motion:

Newton’s 2nd law states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the object’s mass.

Mathematically, we express the second law of motion as follows:

Newton’s Third Law of Motion:

Newton’s 3rd law states that there is an equal and opposite reaction for every action.