A block with mass m = 2 kg is placed against a spring on a frictionless incline

Question

A block with mass m = 2 kg is placed against a spring on a frictionless
incline with an angle ? = 30o. (The block is not attached to the spring). The spring, with a
spring constant k = 19.6 N/m, is compressed 20 cm and then released,
a) What is the elastic potential energy of the compressed spring?
b) What is the change in the gravitational potential energy of the block-Earth system as
the block moves from the release point to its highest point on the incline?
c) How far along the incline is the highest point
from the release point?

Explanation / Answer

a. Energy stored in compressed spring U = (1/2) * k * x2 given k = 19.6 N/cm = 19.6 / 10-2 = 1.96 * 103 N/m x = 20 cm = 0.20 m = 0.5 * 19.6 * 103 * 0.202 = 392 J b. Change in p.e. of system = energy stored in the spring = 392 J c. U = m * g * h 392 = 2 * 9.8 * h vertical height h = 392 / 19.6 = 20.0 m distance covered over the incline x = h / sin ? = 20.0 / sin 300 = 40.0 m

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