Question

A crate of mass 0.812 kg is placed on a rough incline of an angle 35.3 degrees. Near the base of the incline is a spring of spring constant 1140 N/m. The mass is pressed against the spring a distance x and released. It moves up the slope 0.169 meters from the compressed position before coming to a stop. If the coefficient of kinetic friction in 0.195, how far (m) was the spring compressed?

The correct answer is 0.0417 but I get .0369.

My work:

d=kx^2m/2mgsin35.3
.169m = ((1140 Nm)(x^2))/(2(.812 kg)(9.8 m/s^2)(.5778)

(.169m)(9.1967) = 1140 N/m(x^2)

x = sqr root .00136
= .0369

Answer 1

Potential energy in spring = (1/2)(1140)x^2
= 570 x^2

Energy change due to increase in altitude
= m g (.169) sin 35.3
= .812 (9.8)(.169) sin 35.3
= .777 Joules

Energy lost to friction = force*.169
= .195 m g cos 35.3 *.169
= .195 (.812)(9.8)(.169 cos 35.3)
= .214 Joules
so
570 x^2 = .777 + .214 = .991
x = .0417 meters

Answer 2

Thank you so much!!!! I get it.