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ankoles [38]
3 years ago
9

Suppose you lived in a pre-industrial society and needed to lift a heavy (20 kg) block a height of 5 m and had two choices for h

ow to accomplish your task: You could use a lever with a short arm of 1 m and a long arm of 10 m. You'd place the block on the end of the short arm and apply force at the very end of the long arm. You could use a ramp with a length of 8 m and place the block on a cart with wheels. Which choice would you make? Be sure to discuss differences between the effort force of the two options in your answer.
Physics
1 answer:
igomit [66]3 years ago
4 0
Let's break the question into two parts:

1) The force needed in Ramp scenario.
2) The effort force needed in the lever scenario.

1. Ramp Scenario: 
In an incline, the only component of cart's weight(mg) that is in the direction of motion is mgsin \alpha. Therefore the effort force in this case must be equal or greater than mgsin \alpha.

Now we need to find \alpha. \alpha is the angle between the incline of the ramp and the ground. 

Since the height is 5m and the length of the ramp is 8m, sin \alpha would be 5/8 or 0.625. Now that you have sin \alpha, mutiple it with mg.

=> m*g*sin \alpha  = 20 * 10 * 5 / 8. (Taking g = 10 m/s² for simplicity) = 125N
Therefore, the minimum Effort force you would require in this case is 125N.

2. Lever Scenario:
Just apply "moment action" in this case, which is:
F_{e}  d_{e}  = F_{r}  d_{r}

F_{e} = ?

F_{r} = mg = 20 * 10 = 200N
d_{e} = 10m
d_{r} = 1m


Plug-in the values in the above equation:
F_{e} = 200/10= 20N


As 20N << 125N, the best choice is to use lever.

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The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass 1.67 \times 10^(-27)kg, charge +e = +1.60 \times 10^(-19) C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50 \times 10^6 m/s. The proton comes momentarily to rest at a distance 5.31 \times 10^(-13) m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 5.31 \times 10^{-13} m apart?

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The given data is as follows.

Mass of proton = 1.67 \times 10^{-27} kg

Charge of proton = 1.6 \times 10^{-19} C

Speed of proton = 2.50 \times 10^{6} m/s

Distance traveled = 5.31 \times 10^{-13} m

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

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where,    \frac{kq_{p}q_{t}}{r} = U = Electric potential energy

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Putting the given values into the above formula as follows.

        U = (\frac{1}{2}m_{p}v^{2}_{p})

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Therefore, we can conclude that the electric potential energy of the proton and nucleus is 5.218 \times 10^{-15} J.

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