Question

At the instant t = 0 a force F = kt (k is a constant) acts on a small body of mass m resting on a smooth horizontal surface. The time, when body leaves the surface is –

• A. mg k sin a
• B. k sin a/mg
• C. mg sin a/k
• D. mg/ k sin a

Answer 1

Answer: (D)

mg/ k sin a

Answer 2

N+F sin a = mg

N = 0 so

kt sin a = mg

t = $\frac { \mathrm { mg } } { \mathrm { k } \sin \alpha }$