F = kx k = stiffness of the spring x = how much the spring is compressed or stretched

Can you use W = fdcos(theta) to find the work of the spring?

No, because x is a varying force, so you cannot use the work done by constant forces.

The stiffer the spring, the...

higher the k value, meaning it's harder to get them to move. F = -kx can be negative, because the restoring force is in an opposite direction to displacement.

Conservation of Mechanical Energy Equation

1/2mvi^2 + 1/2kxi^2 = 1/2mv^2 + 1/2kx^2

Velocity = 0 (all potential) Xmax, Fmax, amax

Vmax (all kinetic) x = 0 F = 0 a = 0

Velocity = 0 (all potential) Xmax, Fmax, amax

Maximum distance from equilibrium, fixed variable while x is varying -> do not confuse!

Doesn't exactly follow SMH, but as long as the angle is less than 15 degrees, it does.

mg Also remember not to confuse height and amplitude

Main equation to be used throughout the chapter

PEgrav + PEi + KEi = PEgravf + PEf + KEf

Major Calculation Tip

Make sure your calculator is in RADIANS when you are using SMH equations and always calculate carefully.

Some things to know about Etotal at an extreme point

Et at an extreme point = KE + PE = 1/2mv^2 + 1/2kx^2 X at extreme point = A, so Et = 1/2kA^2 At max amp, all energy is converted to PE of the string. 1/2(m1+m2)vf = 1/2kA^2

Tip about maximum compression

Both fk and fspr are going to be positive cause the compression of the string is pushing the box upwards and friction prevents block from moving back downhill. Also mgcos and mgsin for x and y might be switched.

Remember! When you're finding the distance the string has compressed after falling from a height, you'll use (h+x) instead of just h.

You can set that equal to 0. Quadratic formula, if needed, is (-b +- (sqrt (b^2 - 4ac)) / 2a

If you add one block to another while it is traveling through an unstrained string, what will happen to the velocity, amplitude, and maximum speed?

You would use the k/m equation to find that the velocity decreases, you would use conservation of momentum to see that the vmax doesn't change, and you would use the new and old velocities to see that the amplitude increases.

Why does the amplitude increase when you add another block to an unstrained mass?

Because you would require more force to move the mass.

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Physics Chapter 10

Question	Answer
Hooke's Law	F = kx k = stiffness of the spring x = how much the spring is compressed or stretched
Is spring force a conservative force?	Yes!
Potential Energy of Spring Equation	PEelastic = 1/2kx^2
Can you use W = fdcos(theta) to find the work of the spring?	No, because x is a varying force, so you cannot use the work done by constant forces.
The stiffer the spring, the...	higher the k value, meaning it's harder to get them to move. F = -kx can be negative, because the restoring force is in an opposite direction to displacement.
Conservation of Mechanical Energy Equation	1/2mvi^2 + 1/2kxi^2 = 1/2mv^2 + 1/2kx^2
Unstrained String	Velocity = 0 (all potential) Xmax, Fmax, amax
Equilibrium String	Vmax (all kinetic) x = 0 F = 0 a = 0
Strained String	Velocity = 0 (all potential) Xmax, Fmax, amax
Amplitude	Maximum distance from equilibrium, fixed variable while x is varying -> do not confuse!
Simple Pendulum	Doesn't exactly follow SMH, but as long as the angle is less than 15 degrees, it does.
At equilibrium, fx =	mg Also remember not to confuse height and amplitude
Main equation to be used throughout the chapter	PEgrav + PEi + KEi = PEgravf + PEf + KEf
Complete Lecture Problem 1	--
Major Calculation Tip	Make sure your calculator is in RADIANS when you are using SMH equations and always calculate carefully.
Complete Lecture Problem 3	--
Complete Recitation Problem 1	--
Complete Recitation Problem 3	--
Some things to know about Etotal at an extreme point	Et at an extreme point = KE + PE = 1/2mv^2 + 1/2kx^2 X at extreme point = A, so Et = 1/2kA^2 At max amp, all energy is converted to PE of the string. 1/2(m1+m2)vf = 1/2kA^2
Tip about maximum compression	Both fk and fspr are going to be positive cause the compression of the string is pushing the box upwards and friction prevents block from moving back downhill. Also mgcos and mgsin for x and y might be switched.
Remember! When you're finding the distance the string has compressed after falling from a height, you'll use (h+x) instead of just h.	You can set that equal to 0. Quadratic formula, if needed, is (-b +- (sqrt (b^2 - 4ac)) / 2a
If you add one block to another while it is traveling through an unstrained string, what will happen to the velocity, amplitude, and maximum speed?	You would use the k/m equation to find that the velocity decreases, you would use conservation of momentum to see that the vmax doesn't change, and you would use the new and old velocities to see that the amplitude increases.
Why does the amplitude increase when you add another block to an unstrained mass?	Because you would require more force to move the mass.

Created by: smurtab

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