[10分] 物理SHM問題

恩樂

2009-06-07 8:36 pm

A 2kg mass attached to a light spring of spring constant 8 N/cm is placed on a rough inclined plane of inclination 30deg. The other end of the spring is fixed at a wall. The friction between the mass and the inclined plane is a constant of 5N
(a) Find the equilibrium extension of the system.

The mass is initially at the equilibrium position. An impulse is given to it such that it starts the motion with a speed 3m/s
(b) Find the masimum extension of the spring in the subsequent motion

回答 (1)

Prof. Physics

2009-06-07 9:55 pm

✔ 最佳答案

a. By Newton's First law of motion, the net force acting on the mass is zero.

So, mgsin@ = kx + f

(2)(10)sin30* = (800)x + 5

Equilibrium extension, x = 0.00625 m = 0.625 cm

b. The maximum extension occurs when the mass reaches the lowest point. Let e be the maximum extension of the spring.

In the view of conservation of energy,

Loss of K.E. + Loss of G.P.E. = Work done against friction + Gain in E.P.E.

1/2 mv2 + mgesin@ = fe + 1/2 k[(e + x)2 - x2]

1/2 (2)(3)2 + (2)(10)esin30* = (5)e + 1/2 (800)[(e + 0.00625)2 - (0.00625)2]

9 + 10e = 5e + 400(e2 + 0.0125e)

400e2 - 9 = 0

e = 0.15 m

So, the maximum extension of the spring is 0.15 + 0.00625 = 0.156 m

參考: Physics king

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