Question

Suppose that a 200g mass is oscillating at the end of a spring upon a horizontal, frictionless surface. The spring has a spring constant of 240 N/m. In order to make it oscillate, it was originally stretched a distance of 12cm (0.12m) from its unstretched position.
b) What maximum speed will the mass reach during its oscillation?
c) How much kinetic energy does the mass have when the spring is displaced 6cm from its equilibrium position?
Potential energy initially stored in the spring is 1.73

Answer 1

Maximum kinetic energy of the string is=(1/2)*k*A^2=(1/2)*m*v^2

Answer 2

v=?

Answer 3

The abive solution is for b)

Answer 4

c)k.E=(1/2)*k*x^2

Answer 5

Feel free to ask for more clarification

Answer 6

I'm just not sure what numbers to put where. I know that K is my spring constant of 240, but thats about it

Answer 7

m=200gm
A=0.12m

Answer 8

m=200gm=0.2kg

Answer 9

Put those value in my first equation
v=?

Answer 10

So for part C, when I am finding how much kinetic energy the mass has when the spring is displaced 6cm from its equilibrium position, I use the kinetic energy formula, KE=1/2 mv^2 but I am confused as to where I put the displacement value of 6 cm and the other numbers in the equation

Answer 11

C) k.E=(1/2)*k*x^2
Here given k=240N/m
x=6cm=0.06m

Answer 12

I get an answer of 0.432 and it is telling me I am incorrect

Answer 13

Did u write ur qiestion correctly?

Answer 14

I thin ur answer is correct according to ur question

Answer 15

I got it. I had to subtract the .432 from the potential energy that was stored in the spring, giving me an answer of 1.29. Thank you so much for your help!

Answer 16

Glad to help u