Question

A constant force $F_1=10\sqrt{3}N$ applied at an angle ${30}^{\circ }$ with horizontal on body $(m=5\text{kg})$ at rest for $8\text{second}$. In this time interval, body moves from position A to B. To move from position B to A, another constant force $F_2=20\text{N}$ is applied on body . If work done by this force is $960\text{J}$, then find the angle at which the force is acting on the body with horizontal.

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is C ${60}^{\circ }$

When $F_1=10\sqrt{3}\text{N}$ is applied,
$F_x=F_1\cos \theta =10\sqrt{3}\times \cos {30}^{\circ }=15\text{N}$
$a_x=F_x/m=\frac{15}{3}=5{\text{m/s}}^2$
$S=ut+\frac{1}{2}a{t}^2(\because u=0)$

Distance travelled by body from A to B is
$\Rightarrow S=\frac{1}{2}\times 3\times 8^2=96\text{m}$

Let $F_2=20\text{N}$ acts on body makes an angle $\theta$ with horizontal
$W=F_{2}S\cos \theta$
$\Rightarrow 960=20\times 96\cos \theta$
$\Rightarrow \cos \theta =\frac{1}{2}$
$\Rightarrow \theta ={60}^{\circ }$