harddd physics problem, can anyone help?
thanks so much to anyone who is able to help , i will choose best answer.
A moving 3.80 kg block collides with a horizontal spring whose spring constant is 391 N/m. The block compresses the spring a maximum distance of 5.50 cm from its rest position. The coefficient of kinetic friction between the block and the horizontal surface is 0.230. What is the work done by the spring in bringing the block to rest?
I keep getting an answer of 0.591, or 1.06. neither of these are correct.
- JimLv 710 years agoFavorite Answer
SPE = 1/2kx²
SPE = (0.5)(391)(0.055)² = 0.59139 J = work done to compress spring
work done by friction = (ff x 0.055)
ff = (0.230)(normal force) = (0.230)(mg) = (0.230)(3.80)(9.81) = 8.57 J
work done by friction in stopping block = (8.57)(0.055) = 0.4716 J
The way to look at this is:
The spring is compressed by the moving block, AND the friction force does work to stop the block. So the spring's work to stop block is reduced by this friction work. The amount it's reduced is computed above, so that the only work the spring has to do in stopping block is:
0.59139 - 0.4716 = 0.120 J ANS