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user100 [1]
3 years ago
12

A mass of 0.4 kg hangs motionless from a vertical spring whose length is 0.95 m and whose unstretched length is 0.65 m. Next the

mass is pulled down to where the spring has a length of 1.20 m and given an initial speed upwards of 1.6 m/s. What is the maximum length of the spring during the motion that follows
Physics
1 answer:
nexus9112 [7]3 years ago
4 0

Answer:

Explanation:

spring constant of spring = mg / x

= .4 x 9.8 / ( .95 - .65 )

=13.07 N / m

energy stored in spring = 1/2 k x²

= .5 x 13.07 x ( 1.2 - .65 )²

= 1.976 J

Let it goes x m beyond its equilibrium position

Total energy at initial point

= 1.976 + 1/2 m v²

= 1.976 + .5 x .4 x 1.6²

= 2.488 J

energy at final point

= mgh + 1/2 k x²

.4 x 9.8 x  ( .55 + x ) + .5 x 13.07 x² = 2.488

6.535 x² + 2.156 + 3.92 x = 2.488

6.535 x² + 3.92 x - .332 = 0

x = .075 m

7.5 cm

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