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Spring constant

  • #1

Homework Statement




A 150g ball is thrown vertically upward from a window 10m above the ground at a speed of 30m/s. At the top of its trajectory, it compresses a spring a distance, y. (spring constant k = 1.5 N/m). If the spring is 45m above the initial starting position of the ball...

A) how high does the ball go above the ground?
B) what is the speed of the ball when it strikes the ground?



Homework Equations



what i am thinking??? use -2g(y-yo) = vf^2-vi^2 then w=kf-ki then w= -1/2 kx^2

The Attempt at a Solution



those equations i used for part one...i am not sure where to go with part b
 

Answers and Replies

  • #2
kuruman
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This is a prime example of conservation of mechanical energy. Say with an equation that the mechanical energy of the ball right after it is launched is the same as the mechanical energy just before it hits the ground. The spring is irrelevant for part (b).
 
  • #3
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Well the spring should compress the ball so that all the kinetic energy is converted to potential energy.
a)
Kinetics Energy = (1/2)mv^2
Potential Energy = mgh

Do an energy balance.
(1/2)mv^2 - mg(h+x) = (1/2)*kx^2

height = 45 meters
gravity = 9.81 meters/seconds^2
mass = 150 grams = 0.15 kg
velocity = 30 m/s

67.5 J - 1.4715(45+x) = (1/2)*1.5*x^2
67.5 J - 66.218 J -1.4715x = .75 x^2
1.282 -.75x^2 -1.4715x = 0
solve using quadratic equation -> x = 0.653

b) Assuming no air resistance or heat loss the velocity downwards should be the same as upwards.
since acceleration is 9.81 m/s^2 we use simple kinematics:

vf^2 = v0^2 + 2AD

A = acceleration = 9.81 m/s^2
D = distance = 10 meters
v0 = 30 m/s
vf = you can solve for this. Just simple plug and chug
 

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