Question:

A force acts for 20 s on a body of mass 20 kg, starting from rest, after which the force ceases and then body describes 50 m in the next 10 s. The value of force will be:

Updated On: Mar 20, 2025
  • 40 N
  • 5 N
  • 20 N
  • 10 N
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The Correct Option is B

Solution and Explanation

The velocity of the body after 20 s of applied force is: \[ S = v t \quad \Rightarrow \quad 50 = v \times 10 \quad \Rightarrow \quad v = 5 \, \text{m/s} \] Using \(v = u + at\), where \(u = 0\): \[ v = a \times 20 \quad \Rightarrow \quad 5 = a \times 20 \quad \Rightarrow \quad a = \frac{1}{4} \, \text{m/s}^2 \] The force is: \[ F = ma = 20 \times \frac{1}{4} = 5 \, \text{N} \]
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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.