Question

9. What is the total kinetic energy (KEtran + KErot) of a solid cylinder with mass M = 2.50 kg and radius R = 0.5 m that rolls without slipping down a 4.50 m high hill, starting from rest? (Here we assume that there is no energy loss due to friction. I = 1 2 MR2 ,ω = v / R )

Answer 1

Answer:

110.25 J

Explanation:

We are given that

Mass,M=2.5 kg

Radius,R=0.5 m

Distance,d=4.5 m

Initial speed,u=0

We have to find the total kinetic energy

According to law of conservation of energy

Total kinetic energy=Potential energy=mgh

Where g=$9.8m/s^2$

Using the formula

Total kinetic energy=$2.5\times 9.8\times 4.5$

Total kinetic energy$=110.25 J$