Posted by **Casey** on Monday, June 6, 2011 at 6:28pm.

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

- Physics Block and ramp -
**Damon**, Monday, June 6, 2011 at 6:42pm
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

- Physics Block and ramp -
**Casey**, Monday, June 6, 2011 at 7:41pm
Thank you so much!!!! I get it.

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