Question

A box slides from rest down a frictionless ramp inclined at 32.0° with respect to the horizontal and is stopped at the bottom of the ramp by a spring with a spring constant of k = 3.00 104 N/m. If the box has a mass of 12.0 kg and slides 3.00 m from the point of release to the point where it comes to rest against the spring, determine the compression of the spring when the box comes to rest.

Answer 1

0.1117 m

Step-by-step explanation:

Given information

m=12 kg, d=3m ,
`\theta=32^(\circ)`,
`k =3.0*10^(4) N/m`

Net work done W = Fd = mgdcos(90-32) and taking g as 9.81

W = 12*9.81*3*sin(32) = 187.1463 J

From the principal of conservation of energy

W = change in potential energy of spring

`W = 0.5kx^(2)`

`187.1463 = 0.5(3.0*10^(4))x^(2)`

`x=\sqrt {\frac {187.1463}{0.5*(3*10^(4))}`

`x=\sqrt {0.012476}=0.111698\approx 0.1117 m`