A spring with a spring constant 2.7 N/cm is compressed 35 cm and released. The 8

Question

A spring with a spring constant 2.7 N/cm is compressed 35 cm and released. The 8 kg mass skids down the frictional incline at a height of 42 cm from the ground and inclined at a 25 degree angle. The bottom of the ramp is 3 cm off from the ground. The acceleration due to gravity is 9.8 m/s^2. The path is frictionless except for a distance of 0.9 m along the incline which has a coefficient of friction mu = 0.3. What is the final velocity v_f of the mass? 0.9542 m/s 1.6255 m/s 2.7108 m/s 5.6891 m/s None of the above

Explanation / Answer

On ramp,

N - m g cos25 = 0

N = m g cos25

fk =uk N = uk m g

Work done by friction = - 9 fk

= - (0.9 x 0.3 x 8 x 9.8 x cos25)

= - 19.1847 J

Work done by gravity = m g h

= 8 x 9.8 x (0.42 - 0.03)

= 30.576J

Work done by spring = k x^2 /2

= (2.7 x 10^2 N/m) (0.35^2 /2 ) = 16.5375 J

Applying work - energy theorem,

Work done by friction + Work done by gravity + Work done by spring = change in KE

- 19.1847 + 30.576 + 16.5375 = 8 v^2 /2 - 0

v = 2.6424 m/s

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