Question

In the figure, a block of mass m = 20 kg is released from rest on a frictionless incline of angle θ = 31°. Below the block is a spring that can be compressed 2.8 cm by a force of 360 N. The block momentarily stops when it compresses the spring by 5.2 cm.?

(a) How far does the block move down the incline from its rest position to this stopping point? (b) What is the speed of the block just as it touches the spring?

Answer 1

a)`x=3,441 meters`
b)`v=5,897 m/s`

Explanation:

`a)"First we calculate the spring constant:"`
`0,28 mm=0,028 m`
`"K is the spring constant"`
`360=K*0,028`
`K=360/(0,028)=12857,14 N/m`
`"we must calculate the potentiel energy on the compressed spring:"`
`E_p=1/2*K*Delta x`
`E_p=1/2*12857,14*0,052^2`
`E_p=34,766 J`
`34,766=m*g*h`
`34,766=20*9,81*h`
`h=(34,766)/(20*9,81)=(34,766)/(19,62)=1,772 m`
`x*sin 31^o=1,772`
`x*0,515=1,772` ; `x=(1,772)/(0,515)=3,441 meters`
b)
`v^2=2*g*h` ;`v^2=2*9,81*1,772`
`v^2=34,777`
`v=sqrt 34,777`
`v=5,897 m/s`