Question

A small block on a frictionless, horizontal surface has a mass of 2.50×10⁻² kg . It is attached to a massless cord passing through a hole in the surface (Figure 1) .The block is originally revolving at a distance of 0.300 mfrom the hole with an angular speed of 2.99 rad/s . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 m. Model the block as a particle.

-Find the change in the Kinetic Energy of the block

Answer 1

given

mass of 2.50 X 10⁻² kg

distance of 0.300 m

angular speed of 2.99 rad/s

radius of the circle in which the block revolves to 0.150 m

the kinetic energy

K₁ = 0.5 I ₁²

= 0.5 X m X r₁² X ₁²

= 0.5 X 0.025 X 0.3² X 2.99²

K₁ = 0.010057 J

₂ = ( 0.3 / 0.15 ) X 2.99

₂ = 5.98 rad/sec

K2 = 0.5 I ₂²

= 0.5 X m X r₂² X ₂²

= 0.5 X 0.025 X 0.15² X 5.98²

K₂ = 0.010057 J

the change in the Kinetic Energy of the block is = 0.010057 - 0.010057

the change in the Kinetic Energy of the block is zero