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?
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
= 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
Get Answers For Free
Most questions answered within 1 hours.