A 0.50 kg block is attached to a spring of spring constant 32 N/m and set into m

Question

A 0.50 kg block is attached to a spring of spring constant 32 N/m and set into motion. The
block oscillates on a horizontal frictionless surface. The amplitude of the motion is 0.10 m.

(a) What is the magnitude of the block’s maximum acceleration?

(b) What is the maximum speed of the block?

(c) What is the total mechanical energy of the system?

(d) When the block is 0.05 m from the equilibrium position, its speed is …
[ ] less than half its maximum speed
[ ] exactly half its maximum speed
[ ] more than half its maximum speed

(e) At one point at which the block is instantaneously at rest, you carefully drop an additional
mass on top of the block – the block then continues to oscillate, just with extra mass.
Increasing the mass like this changes which of the answers to parts (a) – (c) above? Select all
that apply.
[ ] The answer to (a) would now be different.
[ ] The answer to (b) would now be different.
[ ] The answer to (c) would now be different.

Explanation / Answer

a) block's maximum accln is at one of the ends, where kx = ma ,So, a = kx/m
=32 x 0.1/0.5 m/s2 = 6.4 m/s2

b)maximum speed is got as : 1/2 mv2 = 1/2 kx2 , So, v = x(k/m) = 0.8 m/s

c) total mechanical energy of system = 1/2kx2 = 1/2mv2 = 0.5 x 0.5 x 0.82 = 0.16 J

d)when block is 0.05 m away then , 1/2mv2 + 1/2 k(0.05)2 = 1/2 k (0.1)2 So, v = 0.69 m/s

so,

[ ] more than half its maximum speed

e)

[ ] The answer to (a) would now be different.
[ ] The answer to (b) would now be different.

because, mass differs, but total energy is the same.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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