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# A spring with a spring-constant 2.6 N/cm is compressed 24 cm and released.?

A spring with a spring-constant 2.6 N/cm is compressed 24 cm and released. The 9 kg mass skids down the frictional incline of height 43 cm and inclined at a 24◦ angle.The acceleration of gravity is 9.8 m/s^2. The path is frictionless except for a distance of 0.5 m along the incline which has a coeﬃcient of friction of 0.4. What is the ﬁnal velocity vf of the mass?

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A spring with a spring-constant 2.6 N/cm is compressed 24 cm and released. The 9 kg mass skids down the frictional incline of height 43 cm and inclined at a 24◦ angle.The acceleration of gravity is 9.8 m/s^2. The path is frictionless except for a distance of 0.5 m along the incline which has a coeﬃcient of friction of 0.4. What is the ﬁnal velocity vf of the mass?

Work done by spring = ½ * k * d^2

k = 2.6 N/cm = 260 N/m, d = 0.24 m

Work done by spring = ½ * 260 * .24^2 = 7.488 N * m

Height of incline ÷ Length of incline = sin θ

0.43 ÷ Length of incline = sin 24˚

Length of incline = 0.43 ÷ sin 24˚ = 1.057 m

Force parallel = 9 * 9.8 * sin 24˚

Work done by force parallel = Force parallel * Length of incline

Work done by force parallel = 9 * 9.8 * sin 24˚ * 1.057 = 37.92 N * m

Force friction = -0.4 * 9 * 9.8 * cos 24

Work done by force friction = Force friction * distance of friction

Work done by force friction = -0.4 * 9 * 9.8 * cos 24 * 0.5 = -16.115 N * m

The work done the spring and force parallel increase the energy of the object. The work done by the friction force decreases the energy of the object.

Final energy of the object = 7.488 + 37.92 – 16.115 = 29.293 J

This the kinetic energy of the object at the bottom of the incline

½ * 9 * v^2 = 29.293 J

Final velocity = (29.293 ÷ 4.5)^0.5 = 2.55 m/s

OR

½ * 9 * v^2 = 7.488 + (9 * 9.8 * 0.43) – (0.4 * 9 * 9.8 * cos 24 * 0.5)

½ * 9 * v^2 = 7.488 + (9 * 9.8 * 0.43) – (0.4 * 9 * 9.8 * cos 24 * 0.5)

4.5 * v^2 = 7.488 + 37.926 – 16.115 = 29.299

v^2 = 22.299 ÷ 4.5

v = 2.55 m/s

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