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# Hooke s Law and Spring Potential Energy?

m of block = 1 Kg

K = 100 N/m

Box is stressed 25 inches (0.635 m) and then released.

Find:

1) The position of the block when the kinetic energy is equal to 1/4 the mechanical energy.

2) The velocity of the block when the potential energy is equal to zero.

3) The force on the block when the acceleration is equal to 1/2 the maximum acceleration.

It is a problem we were working on in class and I do not really understand it. Can someone please explain this?

### 1 Answer

- civil_av8rLv 73 years agoFavorite Answer
Total Energy = 1/2*k*A^2 = 1/2*100 N/m *(.635 m)^2 = 20.2 J

Energy at any point = total energy = kinetic energy + spring potential energy

KE = 1/2*m*v^2

SPE = 1/2*k*x^2

1) KE = 1/4*Total Energy

TE = 1/4*TE + SPE

SPE = 3/4*TE

1/2*k*x^2 = 3/4*TE

X = sqrt (3/2*TE/k)

X = sqrt (3/2*20.2 J / 100 N/m) = 0.55 m

2) SPE = 0

TE = KE = 1/2*m*v^2

v = sqrt (2*TE/m)

v = sqrt (2*20.2 J / 1 kg) = 6.36 m/s

3) F_max = k*A = m*a_max

a_max = k/m*A

a_max = 100 N/m / 1 kg * 0.635 m = 63.5 m/s^2

a_new = 1/2*a_max = 31.75 m/s^2

F_new = m*a_new

F_new = 1 kg * 31.75 m/s^2 = 31.75 N

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