Question

A boy pulls a toy with a force of 50 N through a string which makes an angle of 30° with the horizontal so as to move the toy by a distance of 1 m horizontally. If the string were inclined making an angle of 45° with the horizontal, how much pull would he apply along the string in order to move it the same distance of 1 m?

Answer 1

Step 1: Given data

Force with which the boy pulls the toy is, $F=50N$

Angle between the string and the horizontal is, $\theta =30°$

Distance through which the toy moves is, $S=1m$

Step 2: Calculating work done

Work done is given as,

$W=FS\cos \theta$

where, $F$ is force, $S$ is distance and $\theta$ is the angle between force and displacement

Substituting the given values in the above equation for work done, we get,

$$\begin{gathered} W=50N\times 1m\times \cos 30° \\ W=50\times \frac{\sqrt{3}}{2} \\ W=25\sqrt{3}J \end{gathered}$$

Step 3: Calculating force when $\theta =45°$

Substituting the $W=25\sqrt{3}J$, $\theta =45°$ and $S=1m$ in the equation for work done, we get,

$$\begin{gathered} W=FS\cos \theta \\ 25\sqrt{3}J=F\times 1m\times \cos 45° \\ F=\frac{25\sqrt{3}}{\cos 45°} \\ F=\frac{25\sqrt{3}}{1/\sqrt{2}} \\ F=25\sqrt{3}\times \sqrt{2} \\ F\approx 61N \end{gathered}$$

Therefore, force applied by the boy will be $61N$.