Physics Question for Next Step/Blue Print Question Bank

[Pre-Med Oso]

Hey everyone, could someone walk me through solving this MCAT physics problem, Im still not sure how they were able to get to the answer.

What minimum coefficient of friction must a rough, flat surface 10 m in length have if it is to stop a sliding cart with a mass of 100 kg and a velocity of 5 m/s?

A.0.063
B.0.13
C.0.25
D.1.25

Answer is B

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[jhmmd]

Did you start with a FBD?

 

[jhmmd]

I tried to start it for you:

I believe coefficient of friction is the tangent to the inclined plane.

 

[AptarePrep]

Hey everyone, could someone walk me through solving this MCAT physics problem, Im still not sure how they were able to get to the answer.

What minimum coefficient of friction must a rough, flat surface 10 m in length have if it is to stop a sliding cart with a mass of 100 kg and a velocity of 5 m/s?

A.0.063
B.0.13
C.0.25
D.1.25

Answer is B

The free body diagram only has friction in the x-coordinate.

The net sum of the forces is:
Fx = ma

The Fx force is friction force.

Fx = (u)(Fn) (where u is coefficient of friction and Fn is normal force, mass x gravity).

Hence:
(u)(Fn) = (m)(a) (we will come back to this)

Well, acceleration is calculated by this kinematic equation:
Vf^2 = Vi^2 +2 a x
0 = 5^2 + 2 (a) (10)
0 = 25 + 20a
a = 1.25

Now go back to the above equation and estimate the rate of gravity to be 10 m/s^2:
(u) (Fn) = (m) (a)

(u) (100 kg)(10 m/s^2) = (100 kg) (1.25 m/s^2)

u = 1.25/10

u = 0.125 or 0.13

 

[AptarePrep]

Alternatively, you can use energy.

We know that change in kinetic energy is work done on the system.

(1/2)(m)(v^2) = W (and work is force through a distance, r)
(1/2)(m)(v^2) = (F)(r)

In this case, F is due to friction and is equal to (u)(Fn) where u is coefficient of friction and Fn is normal force (mass x gravity).

So:

(1/2)(m)(v^2) = (u)(Fn)(r)

Using 10 for 9.8 m/s^2:

(.5)(100 kg)(25 m^2/s^2) = (u) (100 kg)(10 m/s^2)(10 m)
(.5)(25) = 100u
12.5 = 100u
0.125 = u

 

[doctadank]

Alternatively, you can use energy.

We know that change in kinetic energy is work done on the system.

(1/2)(m)(v^2) = W (and work is force through a distance, r)
(1/2)(m)(v^2) = (F)(r)

In this case, F is due to friction and is equal to (u)(Fn) where u is coefficient of friction and Fn is normal force (mass x gravity).

So:

(1/2)(m)(v^2) = (u)(Fn)(r)

Using 10 for 9.8 m/s^2:

(.5)(100 kg)(25 m^2/s^2) = (u) (100 kg)(10 m/s^2)(10 m)
(.5)(25) = 100u
12.5 = 100u
0.125 = u

Similar to him but this is how I did it and would recommend solving it: (g =10, you can round)
W = Fdcos(theta) theta is 0 so W = Fd
This is the work needed to stop the block where f is the force of friction. This work from friction needs to be equal to the kinetic energy of the block
KE = mv^2/2
Last thing before we set the equations equal is to rewrite F as the friction force. Friction is the friction coefficient times normal force (equal to weight)
F = uf*mg
W = KE, so mv^2/2 = uf*mg*d
Mass cancels out so you get
uk = v^2/(2gd)
Plug and chug: 25/200 = .25/2
Didn’t go past this simplification, looked at which option was closest and picked it which is B.

Once you get used to the process it’s really fast. I was able to do this in my head. Hard to see in writing but I promise it’s really simple if you solve before plugging in