A 0.50-kg block is attached to one end of a horizontal spring with a constant k

Question

A 0.50-kg block is attached to one end of a horizontal spring with a constant k = 200 N/m. The block is pulled a distance of 0.40 m away from equilibrium and released so that it slides back and forth along a frictionless surface.

A. What is the maximum speed of the block?

B. What is the speed when the block has moved half of the way back to equilibrium?

C. What is the kinetic, potential, and total energy of the block at the halfway point?

I'm not arriving at the correct answer so I would greatly appreciate all steps in solving be included in your response!

Explanation / Answer

a) maximum speed is vmax = A*w =

A = 0.4 m

w = sqrt(k/m) = sqrt(200/0.5) = 20 rad/sec

vmax = 0.4*20 = 8 m/sec

B) v = w*sqrt(A^2-x^2)

here x = A/2 = 0.4/2 = 0.2

then v = 20*sqrt(0.4^2-0.2^2) = 6.92 m/sec

C) PE = 0.5*k*x^2 = 0.5*200*0.2^2 = 4 J

KE = 0.5*m*v^2 = 0.5*0.5*6.92^2 = 12 J

TE = 0.5*K*A^2 = 0.5*200*0.4^2 = 16 J

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