The 0.5-kg collar C starts from rest at A and slides with negligible friction on the fixed rod in the vertical plane. Determine the velocity v with which the collar strikes end B when acted upon by the 5-N force, which is constant in direction. Neglect the small dimensions of the collar. [Use Principle of energy conservation]

The answer is 2.32 m/s

Explanation:

Solution

Given that:

U a-b = ΔT

Thus

F cos 30° * hₐ - F sin 30° * hₐ + Whₐ = 1/2 m (v²b - v²ₐ)

Now,

vb = √2Fhₐ (cos 30° - sin 30°) + mghₐ / m + v²ₐ

Where

F = This is the force acting on the collar

m = this is the mass on the collar C

g = The acceleration due to gravity

hₐ = The height of the collar at the position A

vb = The velocity of the collar at position B

vₐ = The velocity of the collar at A

So,

We replace 5N for F, 0.2 m for hₐ, 0.5 kg for m, 9.81 m/s for g, 0 for vₐ

Now,

vb = √ 2 * (5 * 0.2 * (cos 30° - sin 30°) + 0.5 * 9.81 * 0.2 / 0.5 + 0

=√2.694/0.5

=2.32 m/s

Hence, the he velocity v with which the collar strikes the end B is 2.32 m/s