Question

A solid sphere is released from rest and is allowed to roll down a board that has one end resting on the floor and is tilted at 30.0° with respect to the horizontal. If the sphere is released from a height of 26.0 cm above the floor, what is the sphere’s speed when it reaches the lower end of the board?

Answer 1

The speed when it reaches the lower end of the board is 1.9m/s

Step-by-step explanation:

We can use here the equation of energy conservation given by,

`mgh = (1)/(2) mv^2+(1)/(2)I\omega^2`

Where,

`I= (2)/(5)mR^2`

`\omega = (v)/(R)`

m= mass of sphere

v=velocity of sphere

Replacing the equation in our first formula we have,

`mgh = (1)/(2)mv^2+(1)/(2)((2)/(5)mR^2)((v)/(R))^2`

`gh = (1)/(2)v^2+(1)/(5)v^2`

`(9.8)(0.26)=(7)/(10)v^2`

`v=√((9.8)(0.26)(10)/7)`

`v=1.9m/s`

Therefore the speed when it reaches the lower end of the board is 1.9m/s