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?
1 answer