Easy MCQ +4 / -1 PYQ · JEE Mains 2023

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:

  1. A 40 N
  2. B 20 N
  3. C 5 N Correct answer
  4. D 10 N

Solution

<p>$m = 20$ kg</p> <p>$t = 20$ sec.</p> <p>Acceleration $= {F \over {20}}$ m/s$^2$</p> <p>$\therefore$ $v = u + at$</p> <p>$v = 0 + \left( {{F \over {20}}} \right)(20)$</p> <p>$= F$ ms$^{-1}$</p> <p>Now for next 10 sec.</p> <p>$S=ut$</p> <p>$50=F(10)$</p> <p>$F=5$</p>

About this question

Subject: Physics · Chapter: Laws of Motion · Topic: Newton's First Law

This question is part of PrepWiser's free JEE Main question bank. 56 more solved questions on Laws of Motion are available — start with the harder ones if your accuracy is >70%.

Drill 25 more like these. Every day. Free.

PrepWiser turns these solved questions into a daily practice loop. Chapter-wise drills, full mocks, AI doubt chat. No auto-renew.

Start free →