# Work done on a mass by an ideal machine

**1. The problem statement, all variables and given/known data**

An engineer that has a mass of 100kg has made an ideal mechanical advantage machine shown in the attachment. Platforms 1 and 2 are attached to the machine. When he steps on platform 1 it goes straight down and platform 2 rises straight up. The maximum weight he can lift using this machine is twice his own. Platform 1 can be lowered to a maximum of 10m. The mechanical advantage of the machine cannot be adjusted.

Assume the mass (m_{2}) is twice the mass of the engineer (m_{1}) and that he gives a slight push down to get the mass moving. Including the push, the work done on the mass as platform 2 rises to its top height of 5m will be equal to:

a) the original potential energy of the engineer

b) the final kinetic energy of the engineer

c) original potential energy of the engineer plus half the final kinetic energy of the engineer

d) original potential energy of the engineer minus the final kinetic energy of the engineer

The answer is C, but I do not understand why my approach does not work and how they reach their conclusion.

**2. Relevant equations**

m_{1} = mass of the engineer h_{1} = height of platform 1 v_{1} = velocity of m1

m_{2} = mass of the block h_{2} = height of platform 2 v_{2} = velocity of m_{2}

W_{nc} = ΔPE + ΔKE

[tex] \Delta KE = \frac{1}{2} m \Delta v^2[/tex]

ΔPE= mg(h_{f}-h_{o})

**3. The attempt at a solution**

I said that the work done on the m_{2} is

W = (1/2m_{2}v_{f2}^{2} – 1/2m_{2}v_{o2}^{2}) + (m_{2}gh_{2f} – m_{2}gh_{2o})

Seeing as that the final velocity would be zero due to it stopping and that its initial height is 0 (0 reference is the ground). The work done on the mass can be simplified to

W = m_{2}gh_{2f} – 1/2m_{2}v_{o2}^{2}

Then I said that: m_{2} = 2m_{1} h_{2} =1/2h_{1} 2v_{2} = v_{1}

How i found the velocities:

Once the engineer pushes the platforms, they move at a constant velocity until they stop. I said that the initial velocity is the moment right after the engineer pushes and the final velocity is when the platforms are at rest. Bellow is how I found the initial velocity of platform 2 in terms of platform 1.

v_{2} = h_{2}/t

v_{1}=h_{1}/t

h_{1}=2h_{2}

v_{1} = 2h_{2}/t

v_{1}/2 = h_{2}/t

v_{1}/2 = v_{2}

So this leaves the work done on the mass in terms of the potential and kinetic energy of the engineer as:

W = 2m_{1}g(h_{1}/2) – (1/2)(2m_{1})(v_{10}/2)^{2}

W = m_{1}gh_{1} – (1/2)[(1/2)(m_{1}v_{o1}^{2})]

So the work done on the mass is the potential energy from the engineer minus half of the initial kinetic energy of the engineer. This is none of the possible choices and I do not understand were I went wrong. And why my way doesn’t work.

**4. The solution provided**

The net force of both masses is zero so the velocities from the push remain constant. The work done on the mass is W = m_{2}gh_{2f} **+** 1/2m_{2}v_{o2}^{2}, and 2m_{1}=m_{2} 2v_{2}=v_{1} 2h_{2}=h_{1}. The work done on the mass in terms of height, velocity and mass of the engineer is W = m_{1}gh_{1} **+** (1/2)[(1/2)(m_{1}v_{o1}^{2})]

**5. What I do not understand**

How can the final kinetic energy of the engineer not be 0J? The platform is not moving once he moves the maximum distance the platform can travel.

I thought about maybe changing how I define final and initial velocities. I’m thinking maybe I could consider the final velocity to be the velocity right before it stops and the initial velocity would be the platform resting before the push is initiated. If I do that then I get the work done on m_{2} to be:

W = m_{2}gh_{2f} + 1/2m_{2}v_{f2}^{2}.

Once I put it in terms of m_{1} I get W = m_{1}gh_{1o} + 1/2[(1/2)m_{1}v_{f1}^{2}]

If this is the way of getting the answer what was wrong with my original approach to model the problem? Was I not including the push by the engineer?

http://ift.tt/1hztJGR

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