What is the elastic potential energy of a spring with a 80 N/m spring constant that has been moved 0.6 m from equilibrium?

1. F = 0.25N

2. k = 80 N/m

3. x = 0.2m

4. a = 1m/s²

5. x = 42.6m

Explanation:

1. F = kx

 F = 2.5 X 0.1

 F = 0.25N

2. x = 0.25m

  F = 20 N

  F = kx

  20 = k X 0.25

  k = 80 N/m

3. k = 3500 N/m

  F = 700 N

  x = ?

  700 = 3500 x

  x = 0.2m

4. m = 20kg

  k = 244 N/m

  x = 0.1m

  F = 20N

  a = ?

  F = ma

  20 = 20 a

   a = 1m/s²

5. k = 2.3 N/m

  m = 10kg

  x = ?

  F = kx

  10 X 9.8 = 2.3 x

   x = 42.6m

when the velocity of a mass on a spring is zero, the acceleration is at a maximum .

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s