Two crates rest on top of one another.There is a crate of mass 4.62 kg on top of another crate of mass 2.19 kg. The coefficient of friction between the lower crate and the floor is μk = 0.440 and the coefficient of static friction between the two crates is μs = 0.870. If the crates start at rest and a force is applied to the right so that both move a distance of 8.70 m, what is the minimal amount of time required in which this can be accomplished without the top crate sliding on the lower crate? (force applies upper crate) THANK YOU
To find the minimal amount of time required for the crates to move without the top crate sliding on the lower crate, let's break down the problem step by step.
1. Calculate the maximum static friction force between the two crates:
The maximum static friction force (Fs max) can be calculated using the formula: Fs max = μs * Normal force.
The normal force is the force exerted by the lower crate on the upper crate and is given by: Normal force = mass * gravity.
For the top crate: Normal force = (mass of top crate) * gravity = 4.62 kg * 9.8 m/s^2.
For the bottom crate: Normal force = (mass of bottom crate) * gravity = 2.19 kg * 9.8 m/s^2.
2. Calculate the net force required to pull both crates:
The net force needed to overcome static friction (Fs max) between the crates and the force of friction (Ffriction) between the bottom crate and the floor is given by: Net force = Fs max + Ffriction.
The force of friction between the bottom crate and the floor (Ffriction) can be calculated using the formula: Ffriction = μk * Normal force.
3. Calculate the work done by the net force:
The work done by the net force (W) can be calculated using the formula: W = Net force * distance.
The distance is given as 8.70 m.
4. Calculate the power required to do the work in the minimal amount of time:
The power required (P) can be calculated using the formula: P = W / time.
5. Determine the minimal time required:
Rearrange the formula from step 4 to solve for time: time = W / P.
Now, let's calculate the values and find the solution:
1. Calculate the maximum static friction force (Fs max):
Fs max (top crate) = μs * Normal force (top crate).
Fs max (bottom crate) = μs * Normal force (bottom crate).
2. Calculate the net force required to pull both crates:
Net force = Fs max (top crate) + Ffriction (bottom crate).
Ffriction (bottom crate) = μk * Normal force (bottom crate).
3. Calculate the work done by the net force:
W = Net force * distance.
4. Calculate the power required:
P = W / time.
5. Solve for the minimal time:
time = W / P.
By following these steps and substituting the given values into the equations, you can find the minimal amount of time required in which the crates can move without the top crate sliding on the lower crate.