Question

A 0.25 kg mass sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (with k = 800 N/m) whose other end is fixed. The mass has a kinetic energy of 9.0 J as it passes through its equilibrium position (the point at which the spring force is zero).

1.At what rate is the spring doing work on the mass as the mass passes through its equilibrium position?

2.At what rate is the spring doing work on the mass when the spring is compressed 0.075 m and the mass is moving away from the equilibrium position?

Answer 1

Part A.

When mass passes through equilibrium position, then at this point given that spring force will be zero, since

Spring Force = Fs = -k*x

At this point x = 0, So

Fs = 0

Now we need rate at which spring is doing work, So

dW/dt = Power

Power = Force*Velocity = Fs*v

Since Fs = 0 at this point, So

P = rate of work = 0 W

Part B.

Now when spring is compressed by 0.075 m, then

Fs = -k*x = -800*0.075 = -60 N

Now Using energy conservation between equilibrium position and when spring is compressed by 0.075 m

KEi + PEi = KEf + PEf

KEi = 9.0 (given that at equilibrium position KE is 9.0 J)

PEi = 0, (since at this point x = 0)

KEf = (1/2)*m*Vf^2

PEf = (1/2)*k*x^2

m = mass = 0.25 kg

So,

9.0 + 0 = (1/2)*m*Vf^2 + (1/2)*k*x^2

Vf = sqrt [(2*KEi - k*x^2)/m]

Vf = sqrt [(2*9.0 - 800*0.075^2)/0.25]

Vf = 7.35 m/sec

So, Now at this point when spring is compressed 0.075 m, rate of workdone by spring will be:

P = F*v = Fs*Vf

P = -60*7.35

P = -441 W

Please Upvote.