Question

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.

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**

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