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/s2
B
5m/s2
C
1.2m/s2
D
3m/s2

Solution

## 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 $$mgcos30^o=\cfrac{\sqrt3}2mg$$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 $$mgcos60^o=\cfrac12mg$$The frictional force$$=N\times\mu=(\cfrac12mg\times\mu)$$Hence the resultant force acting is $$\cfrac{\sqrt3}2mg-(\cfrac12mg\times\mu)=\cfrac12mg\times(\sqrt3-\mu)$$$$=\cfrac12mg\times(1.732-0.5=1.232)$$Hence the net $$acceleration=\cfrac{force}{mass}$$$$=\cfrac12g\times1.232=\cfrac{9.8}2\times1.232$$It is $$\approx5m/s^2$$Physics

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