In the figure below, a block of mass m = 11 kg is released from rest on a fricti
ID: 1544826 • Letter: I
Question
In the figure below, a block of mass m = 11 kg is released from rest on a frictionless incline angled of angle = 30°. Below the block is a spring that can be compressed 2.0 cm by a force of 265 N. The block momentarily stops when it compresses the spring by 5.3 cm. Set gravitational potential energy of the block-Earth system equal to zero when the block momentarily stops on the compressed spring.
(b) What is the elastic potential energy of the spring when it is compressed 5.3 cm? . J
(c) What is the gravitational potential energy of the block-Earth system when the block is released? J
(d) How far does the block move down the incline from its rest position to its stopping point? . m (
e) What is the kinetic energy of the block just as it touches the spring? J
Explanation / Answer
a)
F = spring force = 265 N
x = compression of spring = 2 cm = 0.02 m
k = spring constant
using the formula
F = kx
265 = k (0.02)
k = 13250 N/m
b)
x' = compression of spring = 0.053 m
elastic potential energy is given as
E = (0.5) k x'2 = (0.5) (13250) (0.053)2 = 18.61 J
c)
using conservation of energy
gravitational potential energy of the block-Earth system when the block is released = E = 18.61 J
d)
m = mass = 11 kg
E = mgh
18.61 = 11 x 9.8 h
h = 0.173 m
L = length of distance travelled on the incline = h/Sin30 = 0.173 /Sin30 = 0.35 m