What's the Force Exerted? and how long it takes to stop

[Hoyin]

Homework Statement

Johnny, of mass 65 kg, and Lucy, of mass 45 kg, are facing each other on roller blades. The coefficient of kinetic friction between the roller blades and concrete surface is 0.20. When Johnny pushes Lucy from rest he applies a force for 1.0 s. Lucy then slows down to a stop in another 8.0 s. Calculate:
a. The applied force exerted by Johnny on Lucy. [790 N]
b. How long it takes Johnny to come to rest. [5.1s]

Homework Equations

F = ma
Vf = Vi + at
x(t) = Vi * t + 0.5*a*t^2

The Attempt at a Solution

Fk = (miu)*mass*9.81 = (0.2) (45Kg) (9.81 m/s^2) = 88.3 N
Vf = Vi + at = 0 + a(8s)

 

[haruspex]

Ouch. This is a very badly worded question. The coefficient of friction between the roller blades and the concrete has nothing to do with it. The roller blades will roll - that is what they are designed to do. The friction will do no work and therefore not slow Lucy at all.
What the question setter means (I hope) is rolling resistance. This is typically a result of friction at the axle of the wheel and/or deformation of the wheel as the load on it changes.
Putting that aside, you have calculated the force that will take Lucy from maximum speed (her speed when the push ends) to rest in 8 seconds. Now you want the force that goes the opposite way - from rest to maximum speed - in 1 second. Any thoughts?
You quote an equation involving an acceleration and two speeds. How can you connect that with forces?

 

[Hoyin]

I tried connecting the two forces Fapp = - F john on lucy but I still need acceleration, which I can't figure out because there is no distance.

 

[haruspex]

I tried connecting the two forces Fapp = - F john on lucy but I still need acceleration, which I can't figure out because there is no distance.

It often happens that it looks like there is not enough information, but it turns out that the information that seems to be missing doesn't affect the answer.
So just create an unknown for the missing data and run through the equations. You'll probably find the unknown cancels out.
In my previous reply I tried to give you a way of thinking about the question which would skip all that, but no matter.

 

[HalfLostAndSearching]

= 8a
x(t) = Vi * t + 0.5*a*t^2 = 0 + 0.5*a*(8s)^2 = 32a

To solve for the applied force exerted by Johnny on Lucy, we can use the equation F = ma. Since we know the mass of Lucy, we can plug in the known values for acceleration and mass to solve for the force. F = (45kg)(8a) = 360a. Since we also know the value for Fk, we can set up the equation F = Fk + ma. Plugging in the known values, we get F = 88.3N + 360a. To solve for a, we can set the two equations equal to each other: 360a = 88.3N + 360a. Solving for a, we get a = 88.3N/360 = 0.2453 m/s^2.

To solve for the time it takes Johnny to come to rest, we can use the equation Vf = Vi + at. Since we know the initial velocity (Vi) is 0, we can plug in the known values for acceleration and final velocity to solve for time. Vf = 0 = 0 + (0.2453 m/s^2)t. Solving for t, we get t = 0.2453 m/s^2/0.2453 m/s^2 = 1s. However, this only represents the time it takes for Johnny to come to rest if he is not applying any force. Since he is pushing Lucy, the force he exerts will cause him to come to rest faster. To calculate the actual time it takes for Johnny to come to rest, we can use the equation x(t) = Vi * t + 0.5*a*t^2. Plugging in the known values, we get x(t) = 0 + 0.5*(0.2453 m/s^2)t^2 = 0.12265t^2. Setting this equation equal to the distance Lucy travels (32a), we get 0.12265t^2 = 32a. Solving for t, we get t = √(32a/0.12265) = √(261.4/0.12265) = 5.1s. This