A block with mass m = 12 kg rests on a frictionless table and is accelerated by a spring with spring constant k = 4784 N/m after being?
A block with mass m = 12 kg rests on a frictionless table and is accelerated by a spring with spring constant k = 4784 N/m after being compressed a distance x1 = 0.491 m from the spring’s unstretched length. The floor is frictionless except for a rough patch a distance d = 2.9 m long. For this rough path, the coefficient of friction is μk = 0.44.
1)How much work is done by the spring as it accelerates the block?J 2)What is the speed of the block right after it leaves the spring?m/s 3)How much work is done by friction as the block crosses the rough spot?J 4)What is the speed of the block after it passes the rough spot?m/s 5)Instead, the spring is only compressed a distance x2 = 0.14 m before being released.How far into the rough path does the block slide before coming to rest?m 6)What distance does the spring need to be compressed so that the block will just barely make it past the rough patch when released?m 7)If the spring was compressed three times farther and then the block is released, the work done on the block by the spring as it accelerates the block is:the samethree times greaterthree times lessnine times greaternine times less
- NCSLv 72 months agoFavorite Answer
1) energy stored in spring was
U = ½kx² = ½ * 4784N/m * (0.491m)² = 577 J
or 580 J to two significant digits
2) became KE:
577 J = ½mv² = ½ * 12kg * v²
v = 9.8 m/s²
3) W = µmgd = 0.44 * 12kg * 9.8m/s² * 2.9m = 150 J
It's possible you'll have to give that a negative sign. I consider it understood.
4) remaining KE is
KE' = (577 - 150) J = 427 J
427 J = ½ * 12kg * v'²
v' = 8.4 m/s
(8.43 before rounding)
5) Now U = ½ * 4784N/m * (0.14m)² = 47 J
and since friction work is proportional to distance we can shortcut with
d = 2.9m * (47/150) = 0.91 m
6) Now we need
U = ½ * 4784N/m * x² = 150 J
x = 0.25 m
or 25 cm
7) nine times greater since spring work is proportional to distance squared
Hope this helps!