Question

A mass of 2.1 kilograms is placed on a horizontal frictionless surface against an uncompressed spring with spring constant 1151.5 N/m. The inclined portion of the surface makes at an angle of 30 degrees to the horizontal and has a coefficient of kinetic friction of 0.27 with the mass. The mass is pushed against the spring until it is compressed a distance 0.15 and then released. How high (vertically), in meters, does the mass rise from the original height before it stops (momentarily) on the inclined ramp?

Answer 1

Initial energy of the mass = 1/2 kA²
= 0.5 x 1151.5 x (0.15)²
= 12.95 J

Kinetic energy of the mass, 1/2 mv² = 1/2 kA² = 12.95
v² = [2 x 12.95] / 2.1 = 12.3375
v = SQRT[12.3375]
= 3.51 m/s

Net deceleration along the ramp, a = gsin + g cos
= 9.81 x sin(30) + 0.27 x 9.81 x cos(30)
= 7.2 m/s²

Final velocity, u = 0
Using the relation, u² - v² = 2as,
Where s is the distance travelled along the ramp
s = [u² - v²] / 2a
= [0 - 12.3375] / [2 x (- 7.2)]
= 0.857 m

Consider h as the vertical height.
h = s sin
= 0.857 x sin(30)
= 0.429 m