An object with mass 1.5 kg is attached to a spring with spring constant k = 280

Question

An object with mass 1.5 kg is attached to a spring with spring constant k = 280 N/m. When the object is 0.05 m from its equilibrium position, it is moving with a speed of 0.27 m/s. Find (a) the total energy of the system, (b) the amplitude of the oscillation, and (c) the maximum speed of the object.

I know the answer is:

(a) 0.4047 J

(b) 0.0538 m

(c) 0.7346 m/s

but I do not know how to do the problem. This is a trigonometry based physics course so please do not use calculus in your answer. Please show all work. Thank you

Explanation / Answer

a) total energy = KE + PE = 1/2*1.5*0.27^2 + 1/2*280*0.05^2 = 0.4046 J

b) 0.4046 = 1/2*KA^2 ( KE = 0 , => whole energy is converted in to PE )

=> A = sqrt ( 2*0.4046)/(280)) = 0.0538 m

c)

0.4046 = 1/2*mVmax^2 ( PE = 0 , => whole energy is converted in to KE )

=> V = sqrt ( 2*0.4046)/(1.5)) = 0.734483 m/s

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