Question

# A 0.225 kg block attached to a light spring oscillates on a frictionless, horizontal table. The...

A 0.225 kg block attached to a light spring oscillates on a frictionless, horizontal table. The oscillation amplitude is

A = 0.190 m

and the block moves at 3.50 m/s as it passes through equilibrium at

x = 0.

(a) Find the spring constant, k (in N/m).

N/m

(b) Calculate the total energy (in J) of the block-spring system.

J

(c) Find the block's speed (in m/s) when x = A/2

m/s.

#### Homework Answers

Answer #1

given
m = 0.225 kg
A = 0.19 m
v_max = 3.5 m/s

a) we know, v_max = A*w

==> w = v_max/A

= 3.5/0.19

= 18.42 rad/s

we know, w = sqrt(k/m)

w^2 = k/m

==> k = m*w^2

= 0.225*18.42^2

= 76.3 N/m

b) Total energy = (1/2)*m*v_max^2

= (1/2)*0.225*3.5^2

= 1.38 J

c) Apply conservation of energy

(1/2)*m*v^2 + (1/2)*k*x^2 = (1/2)*k*A^2

(1/2)*m*v^2 = (1/2)*k*A^2 - (1/2)*k*x^2

v^2 = (k/m)*(A^2 - x^2)

v = sqrt((k/m)*(A^2 - (A/2)^2)

= w*sqrt(A^2 - A^2/4)

= 18.42*sqrt(0.19^2 - 0.19^2/4)

= 3.03 m/s <<<<<<<<-----------Answer

Know the answer?
Your Answer:

#### Post as a guest

Your Name:

What's your source?

#### Earn Coins

Coins can be redeemed for fabulous gifts.

##### Not the answer you're looking for?
Ask your own homework help question