Question

A block of mass, m = 10 kg, starts at the top of a frictionless track which forms a quarter circle of radius r = 10 m. It is given an initial downward velocity of vi = 10 m/s. What is its velocity at the bottom of the track?

Answer 1

Answer:

v=17.32 m/s

Explanation:

Given that

m= 10 kg

R= 10 m

Initial speed ,u= 10 m/s

lets take final speed at the bottom = v m/s

As we know that

Work done by all the forces = Change in the kinetic energy

$mgR=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2$

$2gR=v^2-u^2$

$v=\sqrt{u^2+2gR}$

Now by putting the values

Take g= 10 m/s²

$v=\sqrt{10^2+2\times 10\times 10}$

v=17.32 m/s

The speed of the car at bottom will be 17.32 m/s