Units
Units are in meters
x = Displacement (Elongation/Compression)
A mass is placed on a spring with a spring constant of 15 N/m and the spring is compressed by .25 m. How much force is applied to this spring?
Fs = kx
Fs = 15 N/m * .25m
Fs = 3.75N
A mass is placed on a spring with a spring constant of 15 N/m and the spring is compressed by .25 m. How much potential energy is stored in this spring?
PEs = 1/2 k x2
PEs = 1/2 15 N/m * (.25m)2
PEs = 7.5 N/m * .0625m2
PEs = .47 J
Which spring has the higher spring coefficient?
A- 10 N to stretch 2 m
B- 2 N to stretch .4 m
C- 36 N to stretch 4.8 m
D- 15 N to stretch .15m
Fs = kx
k = Fs / x
A- 10 N / 2 m = 5 N/m
B- 2 N / .4 m = 5 N/m
C- 36 N / 4.8 m = 7.5 N/m
D- 15N / .15 m = 100 N/m
D- 100 N/m
A ball is resting on a shelf when it falls from onto a spring on the ground and is launched into the air again. The ball has a mass of 5 kg. The shelf is 2.5 m above the ground. The spring the ball falls onto has a coefficient of 45 N/m. If 20% of the total energy of the ball is transformed into internal energy at the bottom of the fall due to air resistance, how much does the spring get compressed when the ball lands on it?
ET = PE + KE + Q
ET = GPE + KE + Q, KE = 0, Qbottom = .2ET
ET = GPE = mgh
GPE = 5 kg (9.8m/s2) (2.5 m)
ET = 122.5 J
122.5 J = PE + Q
122.5 J = PEs + .2ET
122.5 J = 1/2 k x2 + .2ET
122.5 J = [(1/2 * 45 N/m) x2] + .2(122.5 J)
122.5 J = [(22.5 N/m) x2] + 24.5 J)
98 J = x2*(22.5 N/m)
x2 = 4.36 m2
x = 2.09 m
The unit for work or energy
J or Joules
A 1 Joule of energy is required to stretch a spring .2 m from its rest position. What is the force required to stretch this spring?
PEs = 1/2 k x2
1 J = 1/2 k (.2m)2
1 J = k (.04m2)
k = 25 N/m
Fs = kx
Fs = 25 N/m * .2m
Fs = 5 N
A 1 Newton of Force is required to stretch a spring .4 m from its rest position. What is the work required to stretch this spring?
Fs = kx
1N = k * .04m
k = 25 N/m
PEs = 1/2 k x2
PEs = 1/2 25 N/m (.4m)2
PEs = 12.5 N/m (.16 m2)
PEs = 2 J
A .66 m long spring is stretch from its equilibrium position to a length of 1.38 m by a weight. If 192.5 J of energy are stored in the stretched spring, what is the value of the spring constant?
PEs = 1/2 k x2
192.5 J = 1/2 k (1.38 m - .66 m)2
192.5 J = k (.2592 m2)
k = 742.7 N/m
A 678 kg test car is traveling at 10.2 m/s when it drives into a giant spring barrier. What is the spring constant for this spring if the spring is compressed 24.8 m? Ignore friction.
ET = PE + KE + Q
ET = KEbefore = PESafter
1/2 mv2 = 1/2 kx2
1/2 678 kg (10.2 m/s)2 = 1/2 k (24.8 m)2
35,269.56 J = (307.52 m2) k
k = 114.7 N/m
The unit for Fs
N or Newtons
The spring of a toy car is wound by pushing the car backward with an average force of ____ newtons through a distance of .5m. How much force is required to store 7.5 Joules of energy within the car?
PEs = 1/2 kx2
7.5 J = 1/2 k (.5m)2
7.5 J = k * .125m2
k = 60 N/m
Fs = kx
Fs = 60N/m .5m
Fs = 30 N
The spring of a toy car is wound by pushing the car backward with an average force of 30 newtons through a distance of .4167m. How much elastic potential energy is stored within the car?
Fs = kx
30N = k *.4167m
k = 72 N/m
PEs = 1/2 kx2
PEs = 1/2 72 N/m (.4167m)2
PEs = 36 N/m * .1736 m2
PEs = 6.25 J
A weight has a mass of 25 kg and is placed on a stiff spring. The weight briefly causes the spring to compress by .08 m. What is the spring constant for this spring?
Fs = kx , FG = W = mg , FG = Fs
Fs = kx = mg
Fs = k (.08 m) = 25 kg (9.8m/s2) = FG
245 N = k (.08 m)
k = 3062.5 N/m
A pop-up toy is at rest and has a mass of .2 kg and has a spring constant of 150 N/m. A force is applied to the toy to compress the spring .075 m. What is the maximum height the toy will rise when the spring is decompressed, if you ignore air resistance?
ET = PE + KE + Q, Q = 0, KE = 0
ET = PE = PEs
ET = PEs = 1/2 k x2
PEs = 1/2 150 N/m (.075 m)2
PEs = 75 N/m (.005625 m2)
PEs = .422 J = ET
ET = PE + KE = (PEs + GPE) + KE
ET = PE = GPE = mgh
.422 J = .2 kg (9.8 m/s2) h
.422 J = h (1.96 N)
h = .215 m
Equivalent units to (kg*m2) / s2
J or Joules
A .2kg apple falls 4m off a branch onto a spring below it. The spring is compressed .25m from its rest position. If all the total energy of the system remains constant, and no energy is lost to air resistance or friction, how much force was generated when this apple fell to the ground and compressed the spring?
N3 says for every action there is an equal and opposite reaction.
Fg = mg
Fg = .2kg (9.8m/s2)
Fg = 1.96 N
The force of the apple falling is equal to the force of the apple compressing the spring, so
Fg = Fs = 1.96 N
A .2kg apple falls from rest 4m off a branch onto a spring below it. The spring is compressed .25m from its rest position. If all the total energy of the system remains constant, and no energy is lost to air resistance or friction, how much elastic potential energy was stored in the spring when this apple fell to the ground and compressed the spring?
ET = PE + KE + Q
Q = 0
KE = 1/2 m v2, v = 0 so KE = 0
ET = GPE = mgh
GPE = .2kg (9.8m/s2) 4m
GPE = 7.8 J
ET = 7.8 J
ET = PE + KE + Q
ET = PES = 7.8 J
A .2kg apple falls from rest 4m off a branch onto a spring below it. The spring is compressed .25m from its rest position. If all the total energy of the system remains constant, and no energy is lost to air resistance or friction, what is the spring constant of this spring?
ET = PE + KE + Q
Q = 0
KE = 1/2 m v2, v = 0 so KE = 0
ET = GPE = mgh
GPE = .2kg (9.8m/s2) 4m
GPE = 7.8 J
ET = 7.8 J
ET = PE + KE + Q
ET = PES = 7.8 J
PES = 1/2 k x2
7.8 J = 1/2 k (.25m)2
7.8 J = k* .03125m2
k = 250 N/m
A 110 kg person jumps from a 3 m high diving board onto a trampoline so they are traveling 6 m/s when they pass their starting height. The trampoline compresses by 2 m when the diver lands on it and is launched into the air. What is the spring constant for the trampoline if the jumper is reaches a height of 4.8 m when they bounce in the air? Ignore air resistance.
ET = PE + KE + Q, Q = 0
ET = KEbefore + GPEbefore = PES = GPEafter
1/2 mv2 + mgh = 1/2 kx2 = mgh
m appears in all equations so we can cancel them out.
1/2 v2 + gh = 1/2 kx2 = gh
1/2 (6 m/s)2 + 10 m/s2 * 3 m = 1/2 k (2 m)2 = 10 m/s2 * 4.8 m
48 J = 2 m2 k = 48 J
k = 24 N/m
The resulting unit AND variable from dividing the Fs by the distance a spring is compress / elongated
N/m AND spring constant
A pop-up toy has a mass of .2 kg and has a spring constant of 150 N/m. A force is applied to the toy to compress the spring .075 m. Calculate the force applied to this toy to compress it.
Fs = kx
Fs = 150 N/m * .075 m
Fs = 11.3 N
A pop-up toy has a mass of .2 kg and has a spring constant of 150 N/m. A force is applied to the toy to compress the spring .075 m. Calculate the potential energy held within the toy.
PEs = 1/2 k x2
PEs = 1/2 150 N/m (.075 m)2
PEs = 75 N/m (.005625 m2)
PEs = .422 J
A collision between a cart and a coil spring results in the spring gaining 320 J of elastic potential energy. The cart comes to rest after compressing the spring .65 m. What is the spring constant for this spring?
PEs = 1/2 k x2
320 J = 1/2 k (.65 m)2
320 J = k (.21125 m2)
k = 1515 N/m
A ball has fallen off a shelf with a speed of 6.7 m/s. When it lands it will fall onto a spring on the ground and will be launched into the air again. The ball has a mass of 7.5 kg. The ball has an additional 1.5 m before it hits the ground. The spring the ball falls onto has a coefficient of 94 N/m. If 12% of the total energy of the ball is transformed into internal energy at the bottom of the fall due to air resistance, how much does the spring get compressed when the ball lands on it?
ET = PE + KE + Q
ET = GPEtop + KEtop + Q: QTop = 0, Qbottom = .12ET
ET = GPE + KE = mgh + 1/2 mv2
ET = mgh + 1/2 mv2 5 kg (9.8m/s2) (2.5 m)
ET = 122.5 J
122.5 J = PE + Q
122.5 J = PEs + .2ET
122.5 J = 1/2 k x2 + .2ET
122.5 J = [(1/2 * 45 N/m) x2] + .2(122.5 J)
122.5 J = [(22.5 N/m) x2] + 24.5 J)
98 J = x2*(22.5 N/m)
x2 = 4.36 m2
x = 2.09 m