Question

A block of mass $m=1kg$ has speed $v=4m/s$ at $\theta ={60}^o$ on a circular track of radiu $R=2m$ as shown in figure. Coefficient of kinetic friction between the block and the track is ${\mu }_k=0.5$. Tangential acceleration of the block at this instant is approximately.

• A. $2.1m/s^2$
• B. $5m/s^2$
• C. $1.2m/s^2$
• D. $3m/s^2$

Answer 1

The correct option is B $5m/s^2$

As shown in the diagram, without friction, the tangential acceleration will be due to gravity only, i.e. $mg.$ Its component along the track is $mgcos{30}^o=\frac{\sqrt{3}}{2}mg$

Now let us consider the friction. It depends on the normal reaction $N$of the track on the block. This reaction is against the component of gravity along it. It is $mgcos{60}^o=\frac{1}{2}mg$

The frictional force$=N\times \mu =(\frac{1}{2}mg\times \mu )$

Hence the resultant force acting is

$\frac{\sqrt{3}}{2}mg-(\frac{1}{2}mg\times \mu )=\frac{1}{2}mg\times (\sqrt{3}-\mu )$

$=\frac{1}{2}mg\times (1.732-0.5=1.232)$

Hence the net $acceleration=\frac{force}{mass}$

$=\frac{1}{2}g\times 1.232=\frac{9.8}{2}\times 1.232$

It is $\approx 5m/s^2$