Question

A 2100 g block is pushed by an external force against a spring (with a 22 N/cm spring constant) until the spring is compressed by 11 cm from its uncompressed length. The compressed spring and block rests at the bottom of an incline of 28◦ with the spring lying along the surface of the ramp.

After all the external forces are removed (so the compressed spring releases the mass) how far D along the plane will the block move before coming to a stop? Answer in units of m.

Answer 1

6.5e-4 m

Step-by-step explanation:

We need to solve this question using law of conservation of energy

Energy at the bottom of the incline= energy at the point where the block will stop

Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy=
`(1)/(2) kx^(2) +PE1`

Energy at the point where the block will stop consists of only gravitational potential energy=
`PE2`

Hence from Energy at the bottom of the incline= energy at the point where the block will stop

⇒
`(1)/(2) kx^(2) +PE1=PE2`

⇒
`PE2-PE1=(1)/(2) kx^(2)`

Also
`PE2-PE2=mgh`

where
`m` is the mass of block

`g` is acceleration due to gravity=9.8 m/s

`h` is the difference in height between two positions

⇒
`mgh=(1)/(2) kx^(2)`

Given m=2100kg

k=22N/cm=2200N/m

x=11cm=0.11 m

∴
`2100*9.8*h=(1)/(2)*2200*0.11^(2)`

⇒
`20580*h=13.31`

⇒
`h=(13.31)/(20580)`

⇒h=0.0006467m=
`6.5e-4`