A 35.8 kg box initially at rest is pushed 2.38 m along a rough, horizontal floor with a constant applied horizontal force of 108.915 N. If the coefficient of friction between box and floor is 0.256, find the work done by the applied force. The acceleration of gravity is 9.8 m/s 2 . Answer in units of J.

The work done by the applied force is 259.22 J.

Explanation:

The work done by the applied force is given by:

Where:

F: is the applied horizontal force = 108.915 N

d: is the distance = 2.38 m  

Hence, the work is:

Therefore, the work done by the applied force is 259.22 J.

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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