Question

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. Force is 8.67N. Also mass displacement is .2m What is the magnitude of the acceleration of the mass when v=v_max/2?

v=.157m/s
KE =0J when x=.1m


Thus wouldn't v_max =2(v) = 2(.157m/s) = .314m/s

i know that F=ma
a=F/m

1/2 mv^2=1/2 k (.1)^2
solve for m/k
m/k=.416

F=kx
8.67N =k(.2m)
k=43.35N/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

Is this correct?

Answer 1

If the kinetic energy is 0 when the displacement is 0.1 m from equilibrium, I don't see how it can get to a displacement of 0.2 m while vibrating on a level table.

At what displacement is the spring force 8.67 N? 0.1m or 0.2m?

The problem makes no sense to me.