Friday
April 29, 2016

# Homework Help: Physics

Posted by Andrea on Sunday, March 2, 2008 at 10:49am.

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=.416

T=2pi*sqrt(m/k)
=2pi*sqrt(.416)
=4.05s

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

Thus wouldn't v_max =2(v) = 2(.157m/s) = .314m/s
V_max=(2pi)/T * x_max
.314m/s=2pi/4.05s * x_max
x_max= .2m

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(.2m)
k=43.35N/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= .416 and k=43.35N/m to find m would I mulitiply .416 by 43.375. This would give me 18kg.

Finally to solve the problem a=8.67N/18kg= .481m/s

Are these correct?
• Physics - drwls, Sunday, March 2, 2008 at 11:07am

a) I get(.1/.157)^2 = .4057 for m/k
b) When the velocity is 1/2 the max, KE is 1/4 of the total, so PE is 3/4 of the total. The spring deflection is therefore sqrt(3/4) of the maximum, or
0.0866 m
Deflection cannot ever be as much as 0.2 m if it stopped at 0.1 m
Make the appropriate corrections and see if you can complete the problem.