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

Question

A disc of mass 3 kg rolls down an inclined plane of height 5 m. The translational kinetic energy of the disc on reaching the bottom of the inclined plane is (A) 50 J (B) 100 J (C) 150 J (D) 175 J

Tardigrade Admin · Accepted Answer

Using mechanical energy conservation m g h=(1/2) m v2+(1/2) l ω2 3(5)(10)=(1/2) m v2+(1/2) m l2((v2/l2)) 150=(3/4) m v2 m v2=200 (1/2) m v2=100 J = K . E. textTranslation

Q. A disc of mass $3 k g$ rolls down an inclined plane of height $5 m$ . The translational kinetic energy of the disc on reaching the bottom of the inclined plane is

50 J

100 J

150 J

175 J

Using mechanical energy conservation
$m g h = \frac{1}{2} m v^{2} + \frac{1}{2} l ω^{2}$
$3 (5) (10) = \frac{1}{2} m v^{2} + \frac{1}{2} m l^{2} (\frac{v ^{2}}{l ^{2}})$
$150 = \frac{3}{4} m v^{2}$
$m v^{2} = 200$
$\frac{1}{2} m v^{2} = 100 J = K . E ._{Translation}$

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