In the figure below, a 2.9 kg block is accelerated from rest by a compressed spr
ID: 1963805 • Letter: I
Question
In the figure below, a 2.9 kg block is accelerated from rest by a compressed spring of spring constant 640 N/m. The block leaves the spring at the spring's relaxed length and then travels over a horizontal floor with a coefficient of kinetic friction µk = 0.25. The frictional force stops the block in distance 6.9 m.http://www.webassign.net/hrw/hrw7_8-51.gif
(a) What is the increase in the thermal kinetic energy of the block floor system?
(b) What was the maximum kinetic energy of the block?
(c) What was the original compression distance of the spring?
Explanation / Answer
a) I would treat the entire problem as a conservation of energy of problem with all the energy lost to friction becoming thermal energy energy. This is just W=Fd where F=mg or 2.9*9.8*.25 times the distance 6.9m which equals 49.02J
b) The lost energy equals the original kinetic so again K=49.02J
c) knowing the kinetic energy, you can set that equal to the potential from the spring so U=.5Kx^2 or 49.02J=.5*640N/m*x^2 or x=.39m. Hope that helped!