A wood crate is sitting on a concrete floor. The crate has a mass of 7.4719x10^2 kg. The crate has an applied force in the x direction and moves with constant velocity in that direction. What is the applied force acting on the crate, to three significant figures, if ìs is 6.678x10^-1 and ìk is 3.752x10^-1?
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at constant velocity, the applied force is equal to friction
frictionforce=mu*weight=mu*mg
To find the applied force acting on the crate, we can use the equation for friction force. The friction force can be calculated using the equation:
Ffriction = ì * Fn
where ì is the coefficient of friction and Fn is the normal force.
In this case, since the crate is sitting on a concrete floor, the normal force is equal to the weight of the crate. The weight of the crate can be calculated using the equation:
Fn = mg
where m is the mass of the crate and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Using the given values, we can calculate the weight of the crate:
Fn = (mass of the crate) * g
= (7.4719x10^2 kg) * (9.8 m/s^2)
Now that we have the normal force, we can calculate the friction force using the given coefficient of friction:
Ffriction = ì * Fn
= (6.678x10^-1) * [(7.4719x10^2 kg) * (9.8 m/s^2)]
Finally, we can find the applied force acting on the crate by considering that the crate is moving with constant velocity, meaning the applied force is equal and opposite to the friction force:
Applied force = -Ffriction
Thus, the applied force acting on the crate is equal to the calculated friction force, to three significant figures.