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 kA^{2}

= 0.5 x 1151.5 x (0.15)^{2}

= 12.95 J

Kinetic energy of the mass, 1/2 mv^{2} = 1/2
kA^{2} = 12.95

v^{2} = [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/s^{2}

Final velocity, u = 0

Using the relation, u^{2} - v^{2} = 2as,

Where s is the distance travelled along the ramp

s = [u^{2} - v^{2}] / 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

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