Posted by
**Andrea** on
.

A block oscillates on a spring and passes its equilibrium position with a speed of .157m/s. It's kinetic energy is zero when the block is at the distance of .1m from equilibrium. Assume no friction between the block and the table.

v=.157m/s

KE =0J when x=.1m

a)First I need to find the period of the oscillation:

1/2 mv^2=1/2 k (.1)^2

solve for m/k

m/k=.405

T=2pi*sqrt(m/k)

=2pi*sqrt(.405)

=4 s

b) what's the mass displacement from the equilibrium when its velocity v=v_mx/2?

x_max=sqrt(3/4)*(.1m)

x_max= .0866m

c) what's the spring constant if the restoring force acting on the mass when its velocity v=v_max/2 is 8.67N?

F=kx

8.67N=k(.0866m)

k=100N/m

d) what's the acceleration of the mass when v=v_max/2?:

i know that F=ma

a=F/m

Given that m/k= .405 and k=100 N/m to find m would I mulitiply .405 by 100 This would give me 40.5kg.

Finally to solve the problem a=8.67N/40.5kg= .21m/s

e)What is the KE of the mass when velocity v= v_max/2?

Isn't KE=1/2mv^2

=1.2(40.5kg)(.157m/s)^2

=.123J

f)What is the PE of the mass when velocity v= v_max/2?

PE= 1/2kx^2

=1/2(100N/m)(.0866m)^2

=.281J

Are these correct?