The GATE Grind

GATE 2017 CE (CE1) – Question 54

Engineering Mechanics · System of forces, free-body diagrams, equilibrium equations · 2 marks · Numerical answer

A 2 kg particle initially travels at 1.5 m/s. A force $f(t)=3t^2$ N acts in its direction of motion for 2 seconds, with $t$ in seconds. Find the velocity immediately after removal of the force in m/s, to one decimal place.

Practise this question in The GATE Grind →

Show answer and explanation

Correct answer: 5.4 to 5.6

Explanation

Impulse is $J=\int_0^2 3t^2dt=[t^3]_0^2=8$ N s. The corresponding velocity increase is $\Delta v=J/m=8/2=4$ m/s. Since the force acts along the initial motion, add this increase: $v=1.5+4=5.5$ m/s. Work–energy could also be used, but would require the displacement history.

More System of forces, free-body diagrams, equilibrium equations questions

All System of forces, free-body diagrams, equilibrium equations questions →