Question

A block is placed at the top of a frictionless inclined plane with angle 30 degrees and released. The incline has a height of 15 m and the block has mass 2 kg. In the above problem the speed of the block when it reaches the bottom of the incline is 8.3 m/s 17 m/s 25 m/s 30 m/s

Answer 1

Answer:17.15 m/s

Step-by-step explanation:

angle of inclination
`(\theta )=30^(\circ)`

height=15 m

mass of block=2 kg

Conserving energy at top and bottom

energy at top of inclined plane(
`E_1`)=mgh

`E_1=2* 9.81* 15=294.3 J`

energy at bottom
`(E_2)=(mv^2)/(2)`

`E_2=(2* v^2)/(2)`

`E_1=E_2`

`294.3=(2* v^2)/(2)`

`v=\sqrt{(2* 294.3)/(2)}`

v=17.15 m/s