Question

A smooth body is released from rest A at a point A at the top of a smooth curved track of vertical height 40 cm, what is the speed of the body at d bottom of the curved track. How far along the adjoining smooth inclined plane will d body go?

Answer 1

Answer:

Explanation:

=> In the absence of friction, the energy of the body is conserved so that on each point of the track we have:

mv²/2 + mgH = constant

0 + mgh = 1/2 mv² + mg

1/2 mv² = mg (40 - 1)

v² = 39 * g * 2

v =  √76440

v = 276.4 cm/s

v = 2.8 m/s

Now, v² = u² + 2gs

v² = u² + 2g( sin 30°) s

Final velocity will be zero here

2.8 * 2.8 = -2 * 9.8 * s / 2

=> neglect negative sign, sice distance can not be negative.

s = 0.8m = 80cm

Answer 2

Answer:

Answer is 2.8 m/s, 80cm

I hope it helps you.