A moving 1.60 kg block collides with a horizontal spring whose spring constant is 295 N/m. The block compresses the spring a maximum distance of 3.50 cm from its rest position. The coefficient of kinetic friction between the block and the horizontal surface is 0.500. What is the work done by the spring in bringing the block to rest?
How much mechanical energy is being dissipated by the force of friction while the block is being brought to rest by the spring?
What is the speed of the block when it hits the spring?
physics again it didnt work. - drwls, Saturday, March 1, 2008 at 12:44pm
I will assume that motion stops in the 3.50 cm compressed position.
The work done by the spring up to that point equals its stored potential energy,
(-1/2) k X^2 = (1/2)*295*(0.035)^2
= -0.1807 J (The spring does negative work while compressing)
The mechanical energy dissipated as friction at that point is the friction force M*g*uk multiplied by the 0.036 m distace moved, or 0.2744 J
For the final part, set the initial kinetic energy equal to the friction work plus the spring compression work (done ON the spring).
(1/2) M V^2 = 0.2744 + 0.1807 = 0.4551 J
V = 0.754 m/s
physics again it didnt work. - rory, Saturday, March 1, 2008 at 3:53pm
get it right thanks
physics again it didnt work. - drwls, Saturday, March 1, 2008 at 4:53pm
get it right or got it right?