JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
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 \(..........\,N\)
- A \(40\)
- B \(5\)
- C \(20\)
- D \(10\)
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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