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Suppose you have a 105-kg wooden crate resting on a wood floor. The coefficients of static and kinetic friction here are fₛ = 0.5 and

fₖ= 0.3. m = 105 kg.If you continue to exert this force once the crate starts to slip, what will its acceleration be, in meters per square second?

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Final answer:

The acceleration of the crate, once it starts to slip, will be -2.94 m/s², which indicates that the crate will slow down due to the kinetic friction force.

Step-by-step explanation:

To calculate the acceleration of a 105-kg wooden crate moving on a wood floor, we need to consider the force of kinetic friction which will act opposite to the motion. Given the mass m = 105 kg and the coefficient of kinetic friction fk = 0.3, we first determine the normal force, which is equal to the weight of the crate. The weight W is calculated by W = mg where g is the acceleration due to gravity (9.80 m/s2). For a 105-kg crate, the normal force N is thus (105 kg)(9.80 m/s2) = 1029 N.

The kinetic friction force Fk is given by Fk = fkN, which is (0.3)(1029 N) = 308.7 N. To find the acceleration a, we use Newton's second law, F = ma, where F is the net force acting on the crate. Since the only horizontal force acting once the crate is moving is the friction force (already calculated), the net force is -308.7 N (negative to indicate opposite direction to the assumed positive direction of motion).

Therefore, a = F/m = -308.7 N / 105 kg = -2.94 m/s2. This negative acceleration indicates that the crate is slowing down due to the kinetic friction force.

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