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Using Energy Equations to find Final Velocity

  • Thread starter Foopyblue
  • Start date
21
1
1. Homework Statement
A box slides down an inclined plane (angle = 37*). The mass of the block is 35 kg, the coefficient of kinetic friction between the box and the ramp is 0.3, and the length of the ramp is 8 m. If it starts from rest at the top of the ramp, with what speed does it reach the bottom? Use energy equations.

2. Homework Equations

[itex]Work = \Delta K [/itex]
[itex]Work = \Delta E [/itex]
[itex]Work = \Delta K + \Delta U_g + \Delta E_t{}_h[/itex]

3. The Attempt at a Solution

I understand how to do this with motion equations:

VF2 = Vi2 +2ad
VF2 = 2ad

a = gsin(θ)-(FFR/m)
a = gsin(37) - (350cos(37)*0.3/35)
a = 6-2.4

VF2 = 2 * 3.6 * 8
VF2 = 57.6
VF = 7.6
-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
But I am at lost with energy equations

I calculated the Net Work of the system by calculating the Work of each Force and adding them together and I got 1020 J.
I want to plug this into either one of these equations:

[itex]Work = \Delta K [/itex]
1020 = (1/2)mv2

[itex]Work = \Delta K + \Delta U_g + \Delta E_t{}_h[/itex]
1020 = (1/2)mv2 + mgy + FFR * d

But what I don't understand is how the first equation can be equal to the second equation. Wouldn't I get a different answer? Why is the change in kinetic energy alone enough to demonstrate the change in all of the energies?
 
Last edited:

ehild

Homework Helper
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But I am at lost with energy equations

I calculated the Net Work of the system by calculating the Work of each Force and adding them together and I got 1020 J.
I want to plug this into either one of these equations:

[itex]Work = \Delta K [/itex]
1020 = (1/2)mv2

[itex]Work = \Delta K + \Delta U_g + \Delta E_t{}_h[/itex]
1020 = (1/2)mv2 + mgy + FFR * d

But what I don't understand is how the first equation can be equal to the second equation. Wouldn't I get a different answer? Why is the change in kinetic energy alone enough to demonstrate the change in all of the energies?
When using an equation, you need to know what the notations mean in it. The three equations you quoted do not mean the same work. Read the context in the text.
.
And look after Work-Energy Theorem in your notes of book. What does it say?
 
21
1
This is what my textbook says for each equation.

Work=ΔE
The total energy of a system changes by the amount of work done on it.

Another way to right the equation above is by listing the specific types of energies found in the system.

The only energies I found in this system were Kinetic, Potential due to gravity, and Thermal so:

Work=ΔK+ΔUg+ΔEth

Work=ΔK
A system gains or loses kinetic energy by transferring it through work between the environment (forces being introduced into the system) and the system.
 
21
1
So if I chose my system to be just the block, I would only have to worry about Kinetic Energy and this is why Work=ΔK works?
 
21
1
I tried thinking about this more deeply. The Law of Conservation of energy states that the total energy of an isolated system is conserved so that its change is zero. If I choose my system to be the Earth, the inclined plane, and the block, then for this problem, it is indeed isolated. This would mean that no work is done. Thus

Work = ΔE
Work = ΔK + ΔUg + ΔEth
0 = ΔK + ΔUg + ΔEth
0 = (Kf - Ki) + (Ugf - Ugi) + ΔEth
[itex] 0 = (\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2)+(mgh_f-mgh_i)+F_F*d[/itex]
vi=0 and hf=0 so the 2nd term and 3rd term go away
[itex] 0 = \frac{1}{2}mv_f^2-mgh_i+F_F*d[/itex]
Friction force is Normal Force times coefficient of kinetic friction, so
[itex] 0 = \frac{1}{2}mv_f^2-mgh_i+(F_N*\mu_k*d)[/itex]
The initial height can be found with trig functions, hypotenuse is 8, searching for opposite side 8sin(37)= 4.8
Normal force is Wcos(37) W = 35*9.8 = 343 343cos(37)= 274
Coefficient of kinetic friction is 0.3
[itex] 0 = \frac{1}{2}*35*v_f^2-35*9.8*4.8+(274*0.3*8)[/itex]
[itex] 0 = 17.5v_f^2-1646.4+657.6[/itex]
[itex] 0 = 17.5v_f^2-988.8[/itex]
[itex] -17.5v_f^2 = -988.8[/itex]
[itex] v_f^2 = 56.5[/itex]
[itex] v_f = 7.5[/itex]

However a system that contains just the block would not be isolated so I would not be able to use the law of conservation of energy if I decided to work with just the block.
I can use Work=ΔK because the definition says that a system gains or loses kinetic energy by transferring it through work between the environment and the system.
I have a system, the block, that is gaining kinetic energy from the environment, the inclined plane and the earth, so I can use this equation.
There are 3 forces on the block, Gravity, Normal Force, and Friction Force. The Normal Force does no work on the system because it is perpendicular to the displacement. A component of gravity, Wsinθ, does work in the direction of movement, and the friction force does work antiparallel to the blocks movement so

(Wsinθ*d)-(Ffr*d) = Total Work
343*sin37*8-82*8 = Total Work
1650-650 = Total Work
1000 J = Total Work

So now that I know the Total Work done on the block

Work=ΔK
[itex]1000 = \frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2[/itex]
[itex]1000 = \frac{1}{2}*35*v_f^2-0[/itex]
[itex]1000 = 17.5v_f^2[/itex]
[itex]57 = v_f^2[/itex]
[itex]7.6 = v_f[/itex]

My numbers came out right but does this logic sound correct?
 

ehild

Homework Helper
15,363
1,771

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