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An object of mass 1.5 kg is moving forwards along the floor against an applied force of
40.0 N [backwards]. If the coefficient of kinetic friction is 0.25, determine the
acceleration of the object.


Sagot :

Answer:

The acceleration of the object is -29.12 m/s².

Explanation:

The acceleration of the object can be calculated by Newton's second law:

[tex] \Sigma F = ma [/tex]

[tex] - F - F_{\mu} = ma [/tex]                        

[tex] - F - \mu mg = ma [/tex]                        

Where:

F: is the applied force = 40.0 N

μ: is the coefficient of kinetic friction = 0.25

m: is the mass of the object = 1.5 kg

g: is the acceleration due to gravity = 9.81 m/s²

a: is the acceleration =?

[tex] a = \frac{- F - \mu mg}{m} = \frac{-40.0 N - 0.25*1.5 kg*9.81 m/s^{2}}{1.5 kg} = -29.12 m/s^{2} [/tex]

The minus sign is because means that the object is decelerating due to the applied force and the friction.  

Therefore, the acceleration of the object is -29.12 m/s².

I hope it helps you!