Conservation of energy of a block problem

[aeroengphys]

Homework Statement

"A block of mass 3.5kg slides down an inclined plane of length 6m that makes an angle of 60° with the horizontal. The coefficient of kinetic friction between the block and the incline is 0.3. If the block is released from rest at the top of the incline, what is its speed at the bottom. Assume that the acceleration due to gravity is 10m/s².

Homework Equations

PE(top) = KE (bottom) + W(nc)
F=ma

The Attempt at a Solution

mgh = 1/2mv² + Fscosθ
(3.5kg)(10m/s²)(5.2m) = 1/2(3.5kg)(v²) + (F)(6m)(cos60)

...F = ma
...F|| - Ff = ma
...mgsinθ - μmgcosθ = ma
...gsinθ - μgcosθ = a
...(10m/s²)(sin60) - (.3)(10m/s²)(cos60) = a
...7.16m/s² = a

182J = 1.75(v²) + (3.5kg)(7.16m/s²)(6m)(cos60)
v² = 61.04m/s
v = 7.8 m/s

The one thing I wasn't sure of was whether or not I substituted for F correctly in the Fscosθ part. If you could let me know if I did that correctly, I'd appreciate it. Thanks in advance.

 

[Doc Al]

The Attempt at a Solution

mgh = 1/2mv² + Fscosθ
(3.5kg)(10m/s²)(5.2m) = 1/2(3.5kg)(v²) + (F)(6m)(cos60)

Looks like you are applying energy methods--a perfectly good approach. But I don't understand your last term, which should be the work done against friction. What is F equal to? (It's not the net force!) You should be able to figure out the force of friction, evaluate the last term, and then solve for v.

...F = ma
...F|| - Ff = ma
...mgsinθ - μmgcosθ = ma
...gsinθ - μgcosθ = a
...(10m/s²)(sin60) - (.3)(10m/s²)(cos60) = a
...7.16m/s² = a

Here you have gone in a completely different direction. You correctly solved for the acceleration. If you wanted to solve for the final speed, you could treat it as a kinematics problem. For some reason you were able to account for friction here, but not earlier in your energy equation. (Compare the terms that I marked in red.)

182J = 1.75(v²) + (3.5kg)(7.16m/s²)(6m)(cos60)
v² = 61.04m/s
v = 7.8 m/s

For some reason, you have (incorrectly) used the acceleration in your energy equation. This doesn't make sense.

There are two (related) ways to solve this: (1) Energy method--as in your first equation, or (2) Find the acceleration and use kinematics. Use one or the other--or both--but keep them separate.

 

[aeroengphys]

I see what you mean. So let's see, if I were to use what i previously solved (acceleration) and were to do this problem using kinematics, I'd get:

x = v0t + ½ at²
(6m) = 1/2(7.16m/s²)t²
t = 1.29s

then:

v = vo + at
v = 0m/s + (7.16m/s²)(1.29s)
v = 9.2m/s

 

[Doc Al]

Good! Now I strongly recommend that you go back and correct your other solution using energy conservation. See if you get the same answer.

 

[aeroengphys]

using the energy method:

mgh = 1/2mv² + Fscosθ
(3.5kg)(10m/s²)(5.2m) = 1/2(3.5kg)(v²) + ((.3)(3.5kg)(10m/s²)(cos60))(6m)(cos0)

work that out and...

v = 9.2 m/s

Thanks for you help Doc Al. We're just beginning conservation of energy, so this is still a little bit new to me. But thanks for helping me point out my mistakes. It seems like my mistakes have helped me learn a new strategy for checking my work...thanks again.