Question

A truck accelerates with $12m/s^2$. A bead can slide frictionlessly on a wire of length $\frac{36}{5}m$ fixed on a truck as shown in the figure. At an instant the bead is released from the lowest point of the wire. Find the time (in second) taken by it to reach the top of the wire.

Answer 1

Draw the diagram showing the direction of different forces.

Find the acceleration $\left(a\right)$ of the bead.

As per the given diagram,

$ma=m{a}^{\prime }\cos {37}^{\circ }-mg\sin {37}^{\circ }$
Therefore,

$a=12\cos {37}^{\circ }-10\sin {37}^{\circ }\dots \dots \left(i\right)$

As we know,

$\cos {37}^{\circ }\frac{4}{5},\sin {37}^{\circ }=\frac{3}{5}$

Hence, by putting the values in equation $\left(i\right)$ we get,

$a=\frac{48}{5}-305=185m/s^2$

Find the time $\left(t\right)$ taken by the bead to reach to the top of the wire.

By using 2nd equation of motion,

$s=ut+\frac{1}{2}a{t}^2\dots \dots \left(ii\right)$

Where, $u=0m/s,s=\frac{36}{5}m$

Therefore, by putting the values in equation $\left(ii\right)$ we get,

$\frac{1}{2}\times \frac{18}{5}\times t^2=\frac{36}{5}$

$t=2s$