Question

A disc of mass 3kg rolls down an inclined plane of height 5m. The transitional kinetic energy of the disc on reaching the bottom of the inclined plane is

Answer 1

h = height of the inclined plane = 5 m

m = mass of the disc = 3 kg

r = radius of the disc

I = moment of inertia of disc

moment of inertia of the disc is given as

I = (0.5) mr²

w = angular speed of the disc

Using conservation of energy

rotational kinetic energy at the bottom + translational kinetic energy at the bottom = Potential energy of disc at top of inclined plane

(0.5) Iw² + (0.5) mv² = mgh

(0.5) (0.5) (mr²) (v/r)² + (0.5) mv² = mgh

(0.25) mv² + (0.50) mv² = mgh

(0.75)  mv² = mgh

mv² = mgh/(0.75)

multiplying both side by (0.5)

(0.5) mv² = (0.5) mgh/(0.75)

Translational KE = (0.5) mgh/(0.75)

Translational KE = (0.5) (3) (9.8) (5)/(0.75)

Translational KE = 98 J

Answer 2

Answer:

By law of conservation of energy

mgh= 3/2 ktrans

Ktrans= 2/3 mgh

= 2/3×3×10× 5

=100J

Explanation: