Question

A body of mass 2kg moves down the quadrant of a circle of radius 4m.The velocity on reaching the lowest point 8ms^-1.Then the loss of energy is

Answer 1

Loss of energy is 14.4 J

Explanation:

Given,

Mass, m = 2 kg

Radius,r = 4 m

Velocity , u = 8 $ms^{-1}$

We know that,

Loss of energy,

= $mgr - \frac{1}{2}\times m\times u^2$

= $2\times 9.8\times 4 - \frac{1}{2}\times 2\times 8^2$

=  $78.4 - 64$

= 14.4 J

Answer 2

The loss in energy is 92.8 J

Explanation:

It is given that,

Mass of the body, m = 2 kg

Radius of circle, r = 4 m

Velocity at the lowest point, v = 8 m/s

Potential energy with respect to the lowest point is,

$P=mgh$

h = 2r

$P=2\times 9.8\times 8$

P = 156.8 J

Kinetic energy at the loweest point is,

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

$K=\dfrac{1}{2}\times 2\times (8)^2$

K = 64 J

The loss in energy is,

L = P - K

L = 156.8 J - 64 J

L = 92.8 J

So, the loss in energy is 92.8 J. Hence, this is the required solution.

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