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JEE Mains · Physics · STD 11 - Newtons laws of motion
  • JEE Mains 2023
  • Medium
  • Numerical

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 \(..........\,N\)

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Answer & Solution

Correct Answer

(B) \(5\)

Step-by-step Solution

Detailed explanation

\(50= V \times 10\)

\(V =5\,m / s\)

\(V =0+ a \times 20\)

\(5= a \times 20\)

\(a =\frac{1}{4}\,m / s ^2\)

\(F = ma =20 \times \frac{1}{4}=5\,N\)

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