Question

A body weighing 25 N is kept over a smooth inclined plane which makes $30^\circ $ with the horizontal. What is the impulse due to weight of the body during the time interval of $10\,{\rm{s}}$?
(a)$25\,{\rm{N}} \cdot {\rm{s}}$
(b)$125\,{\rm{N}} \cdot {\rm{s}}$
(c)$100\,{\rm{N}} \cdot {\rm{s}}$
(d)$75\,{\rm{N}} \cdot {\rm{s}}$

Answer 1

Hint: Impulse can be defined as the total effect of a specified force that acts for a specific time interval that is equal to the value of the change in the momentum. Mathematically the change in the momentum of a body is equal to the multiplication of the amount of the force and the interval of time for that force to act on the body.

Complete step by step answer:
The amount of the weight of the body is :$W = 25\,{\rm{N}}$ .
The angle of the inclined plane with the horizontal is :$\theta = 30^\circ $
The given time interval is $t = 10\,{\rm{s}}$
The impulse is the effect that will act due to the component of the weight along the inclined plane.
So the component of the weight along the inclined plane is given as follows,
 $W\sin \theta $
Now we can calculate the impulse due to the weight of the body along the inclined plane.
$I = W\sin \theta \times t$
Now we will Substitute all the values in the above expression.
$\begin{array}{l}
I = 25\,{\rm{N}} \times \sin 30^\circ \times 10\,{\rm{s}}\\
\Rightarrow I = 25 \times \dfrac{1}{2} \times 10\\
\Rightarrow I = 125\,{\rm{N}} \cdot {\rm{s}}
\end{array}$
Therefore, the value of the impulse due to the weight is $125\,{\rm{N}} \cdot {\rm{s}}$ and the correct option is ${\rm{option(b)}}$.

Note: In such types of questions first find the component of the force. Always check for unit conversion if there is one. The force has unit Newton and the time has unit seconds in the MKS so the unit of the impulse according to the formula is N-s. In the inclined plane problems the force that acts perpendicular to the plane due to the weight of the body is termed as Normal force and this force is equal to the cos component of the weight.