Question

A 0.500-kg block, starting at rest, slides down a 30.0° incline with static and kinetic friction coefficients of 0.350 and 0.250, respectively. After sliding 77.3 cm along the incline, the block slides across a frictionless horizontal surface and encounters a spring (k = 35.0 N/m).What is the maximum compression of the spring?

Answer 1

Answer:x=23.4 cm

Step-by-step explanation:

Given

mass of block
`m=0.5 kg`

inclination
`\theta =30`

coefficient of static friction
`\mu =0.35`

coefficient of kinetic friction
`\mu _k=0.25`

distance traveled
`d=77.3 cm`

spring constant
`k=35 N/m`

work done by gravity+work done by friction=Energy stored in Spring

`mg\sin \theta d-\mu _kmg\cos \theta d=(kx^2)/(2)`

`mgd\left ( \sin \theta -\mu _k\cos \theta \right )=(kx^2)/(2)`

`0.5* 9.8* 0.773\left ( \sin 30-0.25\cos 30\right )=(35* x^2)/(2)`

`x=\sqrt{(2* 0.5* 9.8* 0.773(\sin 30-0.25* \cos 30))/(35)}`

`x=0.234 m`

`x=23.4 cm`