Question

Please show all work

1. A 30 cm spring is compressed to 2 cm. a force of 200 N was used to do this. How much potential energy is stored in the spring?

2. A 5 kg mass is suspended from a spring. The spring develops 100 J of potential energy. What is the spring constant of the spring?

3. A ball is at rest on top of a 15 meter high hill. It rolls down the hill and is stopped by compressing a spring. If the spring is compressed 56 cm, find the spring constant.

4. A 7 kg rock is placed on a spring that has been compressed 48 cm. The spring is released and the rock flies straight upward. As soon as the rock leaves the spring its velocity is 50 m/s. What is the spring constant for the spring?

5. A spring has been stretched 16 cm. The force pulling the spring is reduced so that the stretched length is only 4 cm. By what factor has the potential energy in the spring been reduced?

6. 600N is used to compress a spring to half its initial resting length. How much will the spring be compressed if 800N is used to compress it?

7. 40N stretches a spring from 15 cm to 38 cm. What is the spring constant of the spring?

Answer 1

Answer 2

To compress a spring by a distance $$\Delta$$ x we must apply a force F _ext = k $$\Delta$$ x . By Newton's 3rd law, if we hold a spring in a compressed position, the spring exerts a force F_s = - k $$\Delta$$ x . This is called a linear restoring force because the force is always in the opposite direction from the displacement.
In this problem signs were assumed so that it gets postive and does not effect the value of spring constant

Answer 3

from hookes law, Fspring=-k*x
the other force is gravity Fgravity=mg
you know Fnet=0 as acceleration is 0

mg-kx=0

you can solve all problems with the same formula

Answer 4

1)

potential energy is stored = Work done by the force = F*d = 200*(0.30-0.02) = 56 J ... (converted cm to m)

2)

for SHM of spring we have

F = kx

Potential energy = 1/2*k*x^2 = 1/2*F*x

now,

Answer 5

1. force = kx

200 = k*28/100 (x = 30-2=28)

k= 5000/7

energy = 1/2*k*x^2

= 0.5*(5000/7)*(28/100)^2

=28 joules

2.energy = 1/2*k*x^2

100 = 0.5*k*x^2------(1)

weight = kx

5g = kx-------(2)

dividing both equations give

2 = 0.5*x

x=4

using (2)

50=k*4

k = 12.5 j/m

3.Mass of the block is requires in that case

let mass be m

from conservation of energy

mgh = 1/2*k*(56/100)^2

k is thus obtained

4.from conservation of enrgy

potential energy stored = kinetic energy

1/2*k*x^2 = 1/2*m*v^2

k*(48/100)^2 = 7*50^2

k - 75954.86 j/m

5.Initial potential energy = 1/2*k*16^2

final = 1/2*k*4^2

facotr = initial-final/intial

= (16^2-4^2)/16^2

=0.93

6.let length be l

600 = k*l/2

800=kx

dividing both eqns gives

x = 3l/8

7.F = kx

40 = k(38-15)

k = 40/23 = 1.74