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QuantumEcho wrote:The input and output pistons of a hydraulic jack are respectively 4 cm and 6 cm in diameter. A lever with a mechanical advantage of 9 is used to apply force to the input piston. How much mass can the jack lift if a force of 210 N is applied to the lever?

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Justin72835 wrote:

QuantumEcho wrote:The input and output pistons of a hydraulic jack are respectively 4 cm and 6 cm in diameter. A lever with a mechanical advantage of 9 is used to apply force to the input piston. How much mass can the jack lift if a force of 210 N is applied to the lever?

Hmmm, this reminds me of a similar question on the Mountain View test for this event.

An ME of 9 means that the output force is 9x the input force. This means that the output force from the lever is 1890 N. Now to find the output force of the piston, we can use the relationship F/D^2 = constant to get a force of 840 N. This force is equal to mg, which gives m = 434kg.

Justin72835 wrote:

UTF-8 U+6211 U+662F wrote:

Justin72835 wrote:Next question: You have a solid cube of mass 'm' which is attached to a nearby wall using a massless, ideal spring of constant 'k'. If you launch an arrow of velocity 'v' and mass 'M' directly at the solid cube, what is the maximum compression of the spring if:

1) the arrow sticks into the solid after hitting it?

2) the arrow bounces off of the cube perfectly elastically?

Answer? (Click to Show Hidden Text)

1) kMv/(M+m) 2) 2kMv/(M+m)

You're definitely on the right track (I see that you already found the final velocities of the objects after the collision). Using the final velocity, you can find the kinetic energy of the object after the collision and set it equal to 1/2kx^2 and solve for x. This works because maximum compression occurs when the objects kinetic energy has been converted completely into potential energy.

For those who want it (Click to Show Hidden Text)

[math]x_{inelastic}=Mv\sqrt{\frac{1}{k(M+m)}}[/math]

[math]x_{elastic}=\frac{2Mv}{M+m}\sqrt{\frac{m}{k}}[/math]

1)
[math]Fd = \frac12(M+m)v_{final}^2[/math], where [math]v_{final}[/math] represents the speed of the cube after the collision happens.
[math](kd)d = \frac12(M+m)v_{final}^2[/math]
[math]kd^2 = \frac12(M+m)v_{final}^2[/math]
[math]Mv = (M+m)v_{final}[/math]
[math]v_{final} = \frac{Mv}{M+m}[/math]
[math]kd^2 = \frac12(M+m)\left(\frac{Mv}{M+m}\right)^2[/math]
[math]d^2 = \frac1{2k}(M+m)\left(\frac{Mv}{M+m}\right)^2[/math]
[math]d^2 = \frac1{2k}\frac{(Mv)^2}{M+m}[/math]
[math]d^2 = \frac{(Mv)^2}{2k(M+m)}[/math]
[math]d = \sqrt{\frac{(Mv)^2}{2k(M+m)}}[/math]
[math]d = Mv\sqrt{\frac1{2k(M+m)}}[/math]
But this is slightly different from the posted answer, so where did I go wrong?

2)
[math]kd^2 = \frac12mv_{final}^2[/math]
[math]Mv = Mv_1 + mv_2[/math]
[math]v = v_2 - v_1[/math]
[math]v_1 = v_2 - v[/math]
[math]Mv = M(v_2 - v) + mv_2[/math]
[math]Mv = Mv_2 - Mv + mv_2[/math]
[math]2Mv = v_2(M+m)[/math]
[math]v_2 = \frac{2Mv}{M+m}[/math]
[math]kd^2 = \frac12m(\frac{2Mv}{M+m})^2[/math]
[math]d^2 = \frac1{2k}m(\frac{2Mv}{M+m})^2[/math]
[math]d^2 = \frac{m(2Mv)^2}{2k(M+m)^2}[/math]
[math]d = \frac{2Mv}{M+m}\sqrt{\frac{m}{2k}}[/math]
Again, this answer is slightly different posted answer, so where did I go wrong?