Question

Need a good answer to this question so that i can compare my solutions to yours.

A solid sphere with mass of 4.5 kg rolls without slipping on a horizontal plane with a translational speed of its center of mass of 4.8 m/s. It then rolls up an incline with an angle of 30° .
a) What is its kinetic energy on the horizontal plane?
b) How far does the sphere travel up along the incline?
c) How high would the sphere travel when its mass would be 9.0kg?

Answer 1

a)
Ek=1/2(mv^2)
Ek=1/2(4.5*(4.8)^2)
Ek=51.84J
Ek=52J (2sf)

b) Assuming no friction and g=9.8ms^-2
Considering law of conservation,
Ep=mgh
But we want to find the height, because we already have the Potential energy as found above.
52=4.5*9.8*h
rearranging above
h=52/(4.5*9.8)
h=1.2m (2sf)
now, using trig to calculate the distance up the incline (the hypotenuse)
d=1.2/sin 30
d=2.4m
c) I'm not sure if you meant distance up the incline or the actual vertical height for this question, but you can easily calculate both using the same steps as above, just substituting the mass as 9 instead of 4.5. Then the answer for both vertical height of the ball and the distance up the incline would double.

Hope i have been of some help here. I'm still learning myself and i find helping others (or trying to) is quite good practice.