Acable capable of pulling 4,500 n snapped while trying to drag a 20,000 n compressor across the street. what is the coefficient of static friction for this scenario? note: the 4500 n cable is used as fs and is calculated the same as fk 0.225 > 0.225 4,500 n > 4,500 n 4.44 > 4.44 explain to me how you get your answer.

Explanation:

The cable snapped because the frictional force was greater than the tension in the rope:

Now,

where is the coefficient of static friction, and is the normal force. In our case, the normal force on the compressor is 20,000 N; therefore,

.

Putting this into the condition , we get:

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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