Question

A mass of 5 kg is moving along a circular path of radius 1 m. If the mass moves with 300 rev/min, its kinetic energy would be(a) 250 π²(b) 100 π²(c) 5 π²(d) 0

Answer 1

Pls feel free to ask more doubts

Answer 2

The kinetic energy of the body would be (250 π²) J.

Explanation:

Given that,

Mass of the object, m = 5 kg

Radius of the circular path, r = 1 m

Angular velocity of the object, $\omega=300\ rev/min=31.41\ rad/s$

The linear speed of the body is given by :

$v=r\omega$

$v=1\times 31.41=31.41\ m/s$

The kinetic energy of the body would be :

$K=\dfrac{1}{2}mv^2$

$K=\dfrac{1}{2}\times 5\times (31.41)^2$

$K=2466.4702\ J$

$K=\dfrac{2466.4702}{\pi^2}\times \pi^2$

$K=250\pi^2\ J$

So, the kinetic energy of the body would be (250 π²) J. Hence, this is the required solution.

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