A 2.1×102-kg block on a horizontal frictionless surface is attached to an ideal

Question

A 2.1×102-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 160 N/m. The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the displacement is x = 3.9×102 m, what is the kinetic energy of the block?

I got that the answer 0.512 Joules, but that's not an option.

The options are:

0.37

0.41

0.39

0.44

0.46

And please explain how you got this answer.

Explanation / Answer

Speed at x = 3.9*10^-2 m is v = w*sqrt(A^2-x^2)

w =sqrt(k/m) = sqrt(160/(2.1*10^-2)) = 87.3 rad/s

v = 87.3*sqrt(0.08^2-0.039^2) = 6.09 m/sec

KE = 0.5*m*v^2 = 0.5*(2.1*10^-2)*6.09^2 =0.39 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.