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. A truck is traveling at a speed of 25.0 m/sec along a level road. A crate is resting on the bed of the truck, and the coefficient of static friction between the crate and the truck bed is 0.650. Determine the shortest distance in which the truck can come to a halt without causing the crate to slip forward relative to the truck.

Respuesta :

Answer:

98.12 m

Explanation:

Initial speed(v) = 25.0 m/s

Coefficient of static friction (u) = 0.650

The minimum value of the coefficient of kinetic friction should be the coefficient of static friction.

Kinetic friction is given as

umg = ma (a is the required acceleration)

We then have

a = -ug

= -(0.65*9.8)

= -6.37 m/s^2

The shortest distance d is obtained by

d = (V^2 - v^2)/ 2a

V = 0 m/s (final velocity)

d = (0 - 25^2) / 2(-6.37)

d = -625/-6.37

d = 98.12 m

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