Question

A smooth track in the form of a quarter-circle of radius 6 m lies in the vertical plane. A ring of weight 4 N moves from $P_{1}and{P}_2$ under the action of forces $F_1,F_2$ and $F_3.$ Force $F_1$ is always towards. $P_2$ and is always 20 N in magnitude; force ${\overrightarrow{F}}_2$ always acts horizontally and is always 30 N in magnitude; force ${\overrightarrow{F}}_3$ always acts tangentially to the track and is of magnitude (15-10s) N, where s is in metre. If the particle has speed 4 m/s at $P_1,$ what will its speed be at $P_2,$?

Answer 1

The work done by ${\overrightarrow{F}}_1$ is
$W_1={\int }_{P_1}^{P_2}{F}_1\cos \theta ds$
From figure; $s=R(\frac{\pi }{2}-2\theta )$
or $ds=\left(6m\right)d(-2\theta )=-12d\theta$
and $F_1=20.$
Hence, $W_1=-240{\int }_{\pi /4}^0\cos \theta d\theta$
$=240\sin \frac{\pi }{4}=120\sqrt{2}J$
The work done by ${\overrightarrow{F}}_3$ is
$W_3=\int F_{3}ds={\int }_0^{6\left(\pi /2\right)}(15-10s)ds$
$={[15s-5s^2]}_0^{3\pi }=-302.8JJ$
To calculate the work done by ${\overrightarrow{F}}_2$ and by W, it is convenient to take the projection of the path in the direction of the force, instead of vice-versa. Thus,
$W_2=F_2\left({\overline{OP}}_2\right)=30\left(6\right)=180J$
$W=(-W)\left(\overline{P_{1}O}\right)=(-4)\left(6\right)=-24J$
The total work done is $W_1+W_3+W_2+W=23J$
Then, by the work-energy principle.
$K_{P_2}-K_{P_1}=23J$
$=\frac{1}{2}\left(\frac{4}{9.8}\right)v_2^2-\frac{1}{2}\left(\frac{4}{9.8}\right){\left(4\right)}^2=23$
$v_2=11.3m/s$