A block of mass m = 0.600 kg is fastened to an unstrained horizontal spring whos

Question

A block of mass m = 0.600 kg is fastened to an unstrained horizontal spring whose spring constant is k = 84.8 N/m. The block is given a displacement of +0.154 m, where the + sign indicates that the displacement is along the +x axis, and then released from rest. (a) What is the force (with sign) that the spring exerts on the block just before the block is released? (b) Find the angular frequency of the resulting oscillatory motion. (c) What is the maximum speed of the block? (d) Determine the magnitude of the maximum acceleration of the block.

Explanation / Answer

a) the force of a spring is:
F=-kx=-84.8(.154)
F=-13.0592N
b)For a spring, angular frequency is:

=(k/m)=(84.8/.6)

=11.8884 Hz

c)The max speed is the product of the angular frequency and amplitude

vmax=x=11.8884*.154

vmax=1.831 m/s

c)The max speed is the product of the angular frequency squared and amplitude

amax=2x=(11.884)2(.154)

amax=21.765 m/s2

Hope that helps

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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