First slide

Newton's laws of motion

Question

A force F = bt (where b is a constant) is applied at an angle to a mass m kept on a smooth horizontal plane. The velocity of mass m at the moment of its breaking off the plane is

Difficult

A

$\frac{m g^{2} \cos α}{2 b \sin^{2} α}$
B

$\frac{g^{2} \cos α}{2 m b \sin^{2} α}$
C

$\frac{m g^{2} \sin^{2} α}{2 b \cos α}$
D

$\frac{b g^{2} \cos α}{2 m \sin α}$

Solution

$\begin{array}{l} b t \cos α = m \frac{d v}{d t} . . . . . . (i) \\ N + b t \sin α = m g . . . . . . . . (i i) \end{array}$
For breaking off N=0
$Therefore from (ii), t = \frac{m g}{b \sin α} = t_{0} (say)$
$From (i), m \int_{0}^{v} d v = (b \cos α) \int_{0}^{t_{0}} t d t$
$\begin{array}{l} m v = (b \cos α) \frac{t_{0}^{2}}{2} = \frac{(b \cos α)}{2} \frac{m^{2} g^{2}}{b^{2} \sin^{2} α} \\ v = \frac{m g^{2} \cos α}{2 b \sin^{2} α} \end{array}$

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