Question

Hey guys i need help on this problem

A spring with a spring constant 73.0 n/m sitting vertical on table. The spring is compress 0.210m and a 0.350kg shoe is place on the spring. When the spring is released the shoe shoots into the air. Neglecting air resistance, how high will the shoe go above where it was on the spring?

Answer 1

F = -kx (from newtons law and hookes law)
F = ma
ma = -kx

a = -kx/m

velocity when it returns back to equilibrium (from constant acceleration equations)

V^2 = 2a

height reached

0 = V^2 + 2a times x

solve for x

Answer 2

the a in the last question is the gravitational constant "g" not the acceleration due to the spring "a", I should have made that clearer

Answer 3

so I need to replace the "a" with g are gravitational right?

Answer 4

so your final equation will be

0 = sqrt(-2kx/m) + 2gx

i forgot to square root the v^2 for the last equation

Answer 5

ffs

0 = sqrt (2k times compression) +2gx

solve for x

Answer 6

so the f=-kx are the same as f=ma or two different problem

Answer 7

equation

Answer 8

Force is the same regardless whats doing it. its always kg m s^-2
so you can always equate them

Answer 9

how do i start off the problem im totally lost lol

Answer 10

The question can be solve by two approaches -
- You can use energy balance (1 step method)
- You can use (vertical) projectile motion (2 step method)

Which one would you prefer ??

Answer 11

projectile motion

Answer 12

Step 1: Find the velocity of the shoe at the point it leaves contact with the spring; or in other words, when the spring returns to its normal length.

Can you find it on your own ??

Answer 13

so what i have is f=73.0 x 0.210m f= 153.3

Answer 14

f=ma
f= 153.3 x 0.350 = 438.0

Answer 15

Have you been taught Work and Energy
or SHM ??

Answer 16

i think work and energy like high school stuff

Answer 17

You have to calculate work done by the spring on the shoe (= change in its KE)
You cannot use the equations of motion as acceleration won't be constant.

Answer 18

so how do I do that can you show me?

Answer 19

Which grade do you study in ??

Answer 20

junior in highschool

Answer 21

Right now; I can only give you the formula for velocity -
v = A*ω
where;
v = velocity (at the point when the spring is in equilibrium)
A (called amplitude) = The distance to which the spring is (initially) compressed or stretched
ω (called angular frequency) = √(k/m)
k = spring constant
m = mass of the object

Now find 'v' in this case.

PS: You don't really have to worry about the names of the symbols; I gave them for general knowledge :D

Answer 22

ohh okay thank you

Answer 23

Then, you can use the theory of projectile motion to find the maximum height.

"...how high will the shoe go above where it was on the spring? "
So, don't forget to add 'A' into the answer you get from projectile.

Answer 24

thank you I understand now

Answer 25

@Anthony91
Consider closing the question..