Question

A block is released from the top of a frictionless incline plane as pictured above. If the total distance travelled by the block is 1.2 m to get to the bottom, calculate how fast it is moving at the bottom using Conservation of Energy.

Answer 1

Complete Question

The diagram for this question is showed on the first uploaded image (reference homework solutions )

Answer:

The velocity at the bottom is
`v = 11.76 \ m/ s`

Step-by-step explanation:

From the question we are told that

The total distance traveled is
`d = 1.2 \ m`

The mass of the block is
`m_b = 0.3 \ kg`

The height of the block from the ground is h = 0.60 m

According the law of energy

`PE = KE`

Where PE is the potential energy which is mathematically represented as

`PE = m * g * h`

substituting values

`PE = 3 * 9.8 * 0.60`

`PE = 17.64 \ J`

So

KE is the kinetic energy at the bottom which is mathematically represented as

`KE = (1)/(2) * m v^2`

So

`(1)/(2) * m* v ^2 = PE`

substituting values

=>
`(1)/(2) * 3 * v ^2 = 17.64`

=>
`v = \sqrt{ ( 17.64)/( 0.5 * 3 ) }`

=>
`v = 11.76 \ m/ s`