One last physics question?
A block weighing 2 kg is pushed a spring with negligible mass and a spring constant of k=350 N/m, compressing it 0.25 m. If the block is released, it moves along a horizontal surface (frictionless) and then up an incline (frictionless) with a slope of 25 def.
a) What is the speed of the block as it slides across the flat surface after the spring?
b) How far can the block travel up the ramp before heading back down?
Thanks all, much appreciated.
- NCSLv 77 months agoFavorite Answer
a) convert spring PE to KE
½kx² = ½mv²
½ * 350N/m * (0.25m)² = ½ * 2kg * v²
v = 3.3 m/s
b) convert SPE to GPE
½kx² = mgh = mgLsinΘ
½ * 350N/m * (0.25m)² = 2kg * 9.8m/s² * L * sin25º
L = 1.32 m
OR you could have converted the KE to GPE.
Hope this helps!
- electron1Lv 77 months ago
I assume the slope is 25˚. The maximum kinetic energy of the block is equal to the work done on the spring.
Work = ½ * 350 * 0.25^2 = 10.9375 N * m
KE = ½ * 2 * v^2
v^2 = 10.9375
v = √10.9375
This is approximately 3.3 m/s. The force that causes the block to decelerate is the component of its weight that is parallel to the inclined plane.
Force parallel = 2 * 9.8 * sin 25 = 19.6 * sin 25
The work done by this force is equal to the block’s maximum kinetic energy. Let d be the distance the block travel up the ramp before heading back down.
Work = 19.6 * sin 25 * d
19.6 * sin 25 * d = 10.9375
d = 10.9375 ÷ 19.6 * sin 25
The distance is approximately 1.3 meters.