Aconstruction worker pushes a crate horizontally on a frictionless floor with a net force of 10\, \text 10n, start text, n, end text for 4.0\,\text m4. 0m4, point, 0, start text, m, end text. how much kinetic energy does the crate gain?

= 40J

Explanation:

The work and energy theorem says that:

where is the work of the force, the final kinetic energy and the initial kinetic energy.

Addittionally, the work of the force is calculate as force multiply by distance and if the crate inittialy is at rest, the initial kinetic energy is zero, so:

where F is the force and d the distance. Then, replacing values, we get:

it means that the system gain 40J of kinetic energy.

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

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