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 \(..........\,N\)
Pick an option to check your answer.
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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