Asolid sphere of mass 2.50 kg and radius 0.120 m is at rest at the top of a ramp inclined 15.0°. it rolls to the bottom without slipping. the upper end of the ramp is 1.20 m higher than the lower end. find the sphere’s total kinetic energy when it reaches the bottom.

KE= 30 J

Explanation:

Given that

m= 2.5 kg

r= 0.12 m

θ = 15°

h= 1.2 m

As we know that solid sphere rolls without slipping.It means that

v= ω r

v=Linear velocity

ω=Angular speed

r=radius

The total kinetic energy KE

Moment of inertia of solid sphere I

Now from energy conservation

Energy at top = Energy at bottom

The potential energy at top = m g h

Potential energy at bottom = 0

Kinetic energy at top = 0

Kinetic energy at bottom = KE

m g h + 0 = 0 + KE

KE= 2.5 x 10 x 1.2                 ( take g =10 m/s²)

KE= 30 J

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

i don't understand is there any other way you can rephrase it plz let me know in the comments and i will do it.

explanation: