# If an object with a mass of #10 kg # is moving on a surface at #4 m/s# and slows to a halt after # 6 s#, what is the friction coefficient of the surface?

The initial kinetic energy is lost against the friction losses.

#{ (1/2mv_0^2 = mg mu Delta x), (Delta x = v_0 t - 1/2(mu g)t^2) :}#

we obtain

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To find the friction coefficient of the surface, you can use the equation:

[ \text{Friction force} = \text{Coefficient of friction} \times \text{Normal force} ]

Since the object comes to a halt, the friction force is equal to the force that caused the object to slow down, which is the force of kinetic friction.

The force of kinetic friction can be calculated using Newton's second law:

[ F_{\text{friction}} = m \times a ]

Where ( m ) is the mass of the object (10 kg) and ( a ) is the acceleration of the object. Since the object slows down to a halt, its acceleration is the deceleration, which can be calculated using the equation:

[ a = \frac{{v_f - v_i}}{{t}} ]

Where ( v_f ) is the final velocity (0 m/s), ( v_i ) is the initial velocity (4 m/s), and ( t ) is the time taken (6 s).

Once you have the force of kinetic friction, you can divide it by the normal force to find the coefficient of friction. The normal force is equal to the weight of the object.

[ \text{Coefficient of friction} = \frac{{F_{\text{friction}}}}{{m \times g}} ]

Where ( g ) is the acceleration due to gravity (9.8 m/s²).

[ F_{\text{friction}} = m \times \frac{{v_f - v_i}}{{t}} ] [ \text{Coefficient of friction} = \frac{{m \times \frac{{v_f - v_i}}{{t}}}}{{m \times g}} ]

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