1. Science that goes boing AP Physics Part 1
energy & springs
Science that goes boing
AP Physics Part 1

2. Elastic Potential Energy
Springs are another great way to store potential energy.
Anything that can be stretched out and wants to “snap back” can function as a spring, storing and releasing energy.
We will deal specifically with “ideal” springs that follow Hooke’s Law: the more you stretch them the more force you need to apply.
Fs = -kx

3. Elastic Potential Energy
Fs = -kx
This is Hooke’s Law: x is displacement (how much the spring has been stretched from its neutral position) and Fs is the force on the spring.
“k” is the spring constant, a measure of how “springy” the spring is. Its unit is N/m.

4. Elastic Potential Energy
If a constant force works over a distance, then the work done is just force times distance; the area under the line on this graph is the work done.
In this case, W = force x distance.

5. Elastic Potential Energy
However! If the force is varying, then the work still equals the area… but you need a more complicated equation.
If the force varies directly with the displacement – like in Hookes law, F=kx – the area is one half the area under the curve, ½ bh.
The base “b” is x, the displacement, and the height “h” is F, the force.
But according to Hooke’s law F = kx, so…

6. Elastic Potential Energy
So our equation for work done in compressing a spring is:
W = ½ kx2
And the potential energy held by the spring is:
Us = ½ kx2
This equation is on your equation sheet.

7. Worksheet 2

8. Problems
A spring has a force constant of k=125N/m. It is compressed a distance of 13cm. (a) What is the force required to do this, and (b) how much potential energy is stored in the spring at that compression?
F = kx = (125N/m)(0.13m) = 16N
Us = ½ (125N/m)(0.13m)2 = 1.1J

9. Problems
A kg block, resting on a frictionless surface is pushed 8.00 cm into a light spring, k = 111 N/m. It is then released. What is the velocity of the block as it just leaves the spring?
Set spring potential energy equal to kinetic energy…
½ kx2 = ½ mv so v = x√(k/m)
v = √(111/0.450) = m/s

10. Problems
A 255 g block is traveling along a smooth surface with a velocity of 12.5 m/s. It runs head on into a spring (k = 125 N/m). How far is the spring compressed?
Set spring potential energy equal to kinetic energy…
½ kx2 = ½ mv so x = v √(m/k)
x = (12.5) √(0.255)/125) = m

11. Problems Set spring potential energy equal to kinetic energy…
A 1.0 kg ball is launched from a spring (k = 135 N/m) that has been compressed a distance of 25 cm. The ball is launched horizontally by the spring, which is 2.0 m above the deck. (a) What is the velocity of the ball just after it leaves the spring? (b) What is the horizontal distance the ball travels before it hits? (c) What is the kinetic energy of the ball just before it hits?
Set spring potential energy equal to kinetic energy…
½ kx2 = ½ mv so v = x√(k/m)
v = 0.25m √(135/1.0kg) = 2.9m/s

12. Problems Time to fall: d = ½ at2 t = √(2d/a) = 0.639s
A 1.0 kg ball is launched from a spring (k = 135 N/m) that has been compressed a distance of 25 cm. The ball is launched horizontally by the spring, which is 2.0 m above the deck. (a) What is the velocity of the ball just after it leaves the spring? (b) What is the horizontal distance the ball travels before it hits? (c) What is the kinetic energy of the ball just before it hits?
Time to fall: d = ½ at2 t = √(2d/a) = s
x = vt = (2.9m/s)(0.639s)
x = m

13. Problems
A 1.0 kg ball is launched from a spring (k = 135 N/m) that has been compressed a distance of 25 cm. The ball is launched horizontally by the spring, which is 2.0 m above the deck. (a) What is the velocity of the ball just after it lea ves the spring? (b) What is the horizontal distance the ball travels before it hits? (c) What is the kinetic energy of the ball just before it hits?
You could calculate the total velocity of the ball just before it hits… but it’s easier to do with energy! The final kinetic energy will equal the starting gravitational potential energy (mgh) plus spring potential energy (½ kx2))
mgy0 + ½ kx02 = K
K = 1.0kg(9.8m/s2)(2.0m) + ½ (135N/m)(0.25m)2 = 24J

14. Problems
A spring is compressed a distance of 12 cm by a 675 g ball. The spring constant is 225 N/m. The ball is on a smooth surface as shown. The spring is released and sets the ball into motion. Find: (a) the speed of the ball when it leaves the spring, (b) the speed of the ball just before it leaves the edge of the table, (c) the kinetic energy of the ball just before it hits the deck, (d) the horizontal distance, x, it travels when it leaves the table until it hits the deck. Assume that the ball rolls down the ramp.

15. Problems Set spring potential energy equal to kinetic energy…
A spring is compressed a distance of 12 cm by a 675 g ball. The spring constant is 225 N/m. The ball is on a smooth surface as shown. The spring is released and sets the ball into motion. Find: (a) the speed of the ball when it leaves the spring, (b) the speed of the ball just before it leaves the edge of the table, (c) the kinetic energy of the ball just before it hits the deck, (d) the horizontal distance, x, it travels when it leaves the table until it hits the deck. Assume that the ball rolls down the ramp.
Set spring potential energy equal to kinetic energy…
½ kx02 = ½ mv2
v = x0√ (k/m) = (0.12m) √(225/.675) = 2.2m/s

16. Problems Kinetic energy plus potential energy equals final K
A spring is compressed a distance of 12 cm by a 675 g ball. The spring constant is 225 N/m. The ball is on a smooth surface as shown. The spring is released and sets the ball into motion. Find: (a) the speed of the ball when it leaves the spring, (b) the speed of the ball just before it leaves the edge of the table, (c) the kinetic energy of the ball just before it hits the deck, (d) the horizontal distance, x, it travels when it leaves the table until it hits the deck. Assume that the ball rolls down the ramp.
Kinetic energy plus potential energy equals final K
(and let’s look just as the first drop, a fall of 1.7 – 1.0 = 0.7m)
mgy0 + ½ mv02 = ½ mv2 v = √(2(½ v02 + gy0))
v = √(2(½(2.2m/s)2 + (9.8)(1.7m – 1.0m)) = 4.3m/s

17. Problems
A spring is compressed a distance of 12 cm by a 675 g ball. The spring constant is 225 N/m. The ball is on a smooth surface as shown. The spring is released and sets the ball into motion. Find: (a) the speed of the ball when it leaves the spring, (b) the speed of the ball just before it leaves the edge of the table, (c) the kinetic energy of the ball just before it hits the deck, (d) the horizontal distance, x, it travels when it leaves the table until it hits the deck. Assume that the ball rolls down the ramp.
Total kinetic energy equals starting spring potential energy plus starting gravitational potential energy
mgy + ½ kx2 = total K
K = (0.675kg)(9.8)(1.7m) + ½(225)(0.12m)2 = 13J

18. Problems Falling time: t = ((2)(1.0m)/(9.8)) = 0.452s
A spring is compressed a distance of 12 cm by a 675 g ball. The spring constant is 225 N/m. The ball is on a smooth surface as shown. The spring is released and sets the ball into motion. Find: (a) the speed of the ball when it leaves the spring, (b) the speed of the ball just before it leaves the edge of the table, (c) the kinetic energy of the ball just before it hits the deck, (d) the horizontal distance, x, it travels when it leaves the table until it hits the deck. Assume that the ball rolls down the ramp.
Falling time: t = ((2)(1.0m)/(9.8)) = s
distance = (4.3m/s)(0.452s) = m

19. Problems
A 49 kg projectile has a kinetic energy of 825 kJ when it is fired at an angle of 23.0. Find (a) the time of flight for the projectile and (b) the maximum range of the projectile.
Step one: find the velocity from the kinetic energy, and then the x and y components of the velocity.
K = ½ mv2 so v = √(2K/m) = 184m/s
vy = (184m.s)sin23.0° = m/s
vx = (184m.s)cos23.0° = 169m/s

20. Problems
A 49 kg projectile has a kinetic energy of 825 kJ when it is fired at an angle of 23.0. Find (a) the time of flight for the projectile and (b) the maximum range of the projectile.
Find the time using the y component of velocity:
t = (vy – vy0)/g = (-71.9 – 71.9)/(9.8) = 15s

21. Problems
A 49 kg projectile has a kinetic energy of 825 kJ when it is fired at an angle of 23.0. Find (a) the time of flight for the projectile and (b) the maximum range of the projectile.
Find the range using the time and the x component of velocity:
x = vt = (169m/s)(15s) = 2,500m

22. Problems
A 49 kg projectile has a kinetic energy of 825 kJ when it is fired at an angle of 23.0. Find (a) the time of flight for the projectile and (b) the maximum range of the projectile.
Find the range using the time and the x component of velocity:
x = vt = (169m/s)(15s) = 2,500m