A 2.00kg object is attached to a spring. The force constant of the spring is 20.

Question

A 2.00kg object is attached to a spring. The force constant of the spring is 20.0 N/m. The object is displaced 0.0500m from the equilibrium position. After release, the object exhibits SHM. Determine:

            A) Period of the motion: ANS (2.81 s)

            B) Total energy of the system: ANS (0.0250 J)

            C) Object’s speed when it is 0.0100 meter from equilibrium: ANS (0.154 m/s)

Physics 1 College level. Please show all work and formula used and that the answer mathes the one given. Thanks!

Explanation / Answer

Given that

mass m=2 kg

spring constant K=20 N/m

amplitude A=0.05 m

now we find the time period of the motion

time period T=2*3.14[m/k]^1/2

=6.28*[2/20]^1/2

=1.99 sec

now we find the total energy of the system

total energy =1/2*k*A^2=1/2*20*0.05^2=0.025 J

now we find the speed

speed V=(20/2)^1/2*(0.05^2-0.01^2]^1/2

=3.2*0.049

=0.154 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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