please help! its due in 30 minutes A 1.5 kg solid sphere (radius = 0.20 m ) is r

Question

please help! its due in 30 minutes A 1.5 kg solid sphere (radius = 0.20 m ) is released from rest at the top of a ramp and allowed to roll without slipping. The ramp is 0.60 m high and 6.0 m long.

PART A:

When the sphere reaches the bottom of the ramp, what is its total kinetic energy?

K=_____J

PART B:

When the sphere reaches the bottom of the ramp, what is its rotational kinetic energy?

Kr=____J

PART C: When the sphere reaches the bottom of the ramp, what is its translational kinetic energy?

Kt= _____J

Explanation / Answer

here,

mass , m = 1.5 kg

radius , r = 0.2 m

h0 = 0.6 m

a)

using conservation of energy

the total kinetic energy gained = potential energy lost

the total kinetic energy = m * g * h0 = 1.5 * 9.81 * 0.6

the total kinetic energy is 8.83 J

b)

let the final speed be v

total kinetic energy = rotational kinetic energy + translational kinetic energy

8.83 = 0.5 * I * w^2 + 0.5 * m * v^2

8.83 = 0.5 * (2/5 * m * r^2) * (v/r)^2 + 0.5 * m * v^2

8.83 = 0.7 * m * v^2

8.83 = 0.7 * 1.5 * v^2

v = 2.9 m/s

so the rotation kinetic energy = 0.2 * m * v^2 = 0.2 * 1.5 * 2.9^2 = 2.52 J

c)

the translational kinetic energy , TE = 0.5 * m * v^2 = 0.5 * 1.5 * 2.9^2

TE = 6.3 J

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