A spring with constant 559 N/m is 0.14 meters long when not stretched or compressed. You compress the spring until it is 0.07 meters long. How many joules of elastic potential energy are stored in the compressed spring

1.46755 Joules

Explanation:

Applying,

E = 1/2ke² Equation 1

Where E = Elastic potential energy of the spring, k = spring constant, e = extension

From the question,

Gieven: k = 559 N/m, e = (0.14-0.07) meters = 0.07 meters

Substitute these values into equation 1

E = 1/2(599×0.07²)

E = 2.9351/2

E = 1.46755 Joules

it doesn't. it was imagined up in the middle ages.

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