Question

A 1.0-kg block is released from rest at the top of a frictionless incline that makes an angle of 37° with the horizontal. An unknown distance down the incline from the point of release, there is a spring with k = 200 N/m. It is observed that the mass is brought momentarily to rest after compressing the spring 0.20 m. How far does the mass slide from the point of release until it is brought momentarily to rest?

Answer 1

Answer: The distance along the incline the mass slide from the point of release until it is brought momentarily to rest is 0.68 meter

Step-by-step explanation:

Let distance slide by block along the incline be L

`\therefore h=L\sin (37^(\circ))`

where h = initial height of block from the ground level

Since the inclined surface is frictionless so total mechanical energy is conserved for the spring-block system

Therefore applying conservation of mechanical energy we get

`PE_(si)+PE_(gi)+KE_i=PE_(sf)+PE_(gf)+KE_f`

where

`PE_(si)` = Initial spring potential energy

`PE_(gi)` = Initial gravitational potential energy

`KE_i` = Initial kinetic energy

`PE_(sf)` = Final spring potential energy

`PE_(gf)` =Final gravitational potential energy

`KE_f` = Final kinetic energy

=>
`0+mgL\sin (37^(\circ))+0=(kx^(2))/(2)+0+0`

=>
`1.0* 9.8* L\sin (37^(\circ))=(200* 0.20^(2))/(2)`

`\therefore L= 0.68 m`

Thus the distance along the incline the mass slide from the point of release until it is brought momentarily to rest is 0.68 meter