potential energy stored in spring = .5 k x^2
= .5 * 4201 * .556^2 = 649 Joules
that becomes Ke of block
.5 m v^2 = 649 Joules
now if that is used up in frriction
mu m g * d = 649.
.49 * 17 * 9.81 * d = 649
solve for d
that d is the distance it can pass over a rough patch.
Now that gives you d for a given x of 0.556 m, you have to reverse that
to solve for x given d = 2.3 m
however to make it easy we know that x^2 is proportional to d
A block with mass m = 17 kg rests on a frictionless table and is accelerated by a spring with spring constant k = 4201 N/m after being compressed a distance x1 = 0.556 m from the spring’s unstretched length. The floor is frictionless except for a rough patch a distance d = 2.3 m long. For this rough path, the coefficient of friction is μk = 0.45.
1) What distance does the spring need to be compressed so that the block will just barely make it past the rough patch when released?
1 answer