Question

QUESTION ABOUT PULLEYS AND CRATES

Answer 1

Ahahaha, let's see.

Answer 2

It says the start at rest, so both will have 0 kinetic energy.

Answer 3

since \(KE = \frac 12 mv^2 = \frac 12 m(0)^2 = 0\)

Answer 4

In both cases, \(PE = mgh\), however we know that for the second crate has \(h=0\) so \(PE = 0\).

For the first crate, we don't know the height, but we do know the \(m\) and \(g\) so we plug them in to be more precise.

Answer 5

im actually on part (d) so...... im lookin for (V)

Answer 6

okay, can you find the potential energy of \(m_1\) then?

Answer 7

Then find the potential energy of \(m_2\) once it reaches the same height.

Answer 8

is there such thing as V= \[\sqrt{gh}\]..like v= \[\sqrt{2gh}\] but without the 2

Answer 9

The difference would have to be the energy that remains in kinetic energy.

Answer 10

Well use \[
KE = \frac 12mv^2\implies v = \sqrt{\frac{2KE}{m}}
\]

Answer 11

m being the total mass?

Answer 12

To get KE, you need to find
Initial total energy, which is PE of \(m_1\).
Final total energy, which is PE of \(m_2\) and KE of \(m_1\).

KE of \(m_1\) is what you use.

Answer 13

.....


Okay, what is the total energy at the start before anything falls?

Answer 14

0?

Answer 15

No, at the start one of the boxes has potential energy.

Answer 16

265 J

Answer 17

PE of m1 is 402

Answer 18

Ok, when m1 is at the ground, what is the PE of m2?

Answer 19

i really don't know.....

Answer 20

Well, you know it will be as high as m1 used to be.

Answer 21

you also know m2 and g... so you can calculate it.

Answer 22

137.34

Answer 23

Okay, so the kinetic energy of m1 is whatever energy remains once you take away the potential energy of m2 from the initial total energy

Answer 24

265?

Answer 25

Yes. That should be the total kinetic energy of both masses.

Answer 26

So now, we use the formula I showed you earlier to find the velocity of the masses

Answer 27

so V= \[\sqrt{2(295}/55 ?\]

Answer 28

yeah, try it see what you get

Answer 29

not 5

Answer 30

Total Energy Before (b) = Total Energy After (a)
$$
PE_1^b+PE_2^b+KE_1^b+KE_2^b=PE_1^a+PE_2^a+KE_1^a+KE_2^a\\
PE_1^b=35\times9.81\times 2.2=755.37~Joules\\
PE_2^b=0~Joules\\
KE_1^b=0~Joules\\
KE_2^b=0~Joules\\
PE_1^a=0~Joules\\
PE_2^a=14\times9.81\times 2.2=302.148~Joules\\
KE_1^a=\cfrac{1}{2}m_1v^2=\cfrac{1}{2}35\times v^2~Joules\\
KE_1^a=\cfrac{1}{2}m_2v^2=\cfrac{1}{2}14\times v^2~Joules\\
$$
So,
$$
PE_1^b+PE_2^b+KE_1^b+KE_2^b=PE_1^a+PE_2^a+KE_1^a+KE_2^a\\
755.37=302.148+\left (\cfrac{35}{2}+7\right )v^2\\
\implies v=\sqrt{\cfrac{755.37-302.148}{\cfrac{35}{2}+7}}=4.30~m/s
$$
Hope this makes sense.

Answer 31

nevermind...... their was a confusion between two similar questions I had, but yes, 4.30 is the answer....


THANK YOU!!! @ybarrap

Answer 32

Here is a way to check that our anwer of 4.30 m/s is correct.

Summing forces in the y-direction for each FBD we get
$$
35g-T=35a\\
T-14g=14a
$$
Once you get a, the acceleration of the blocks, use
$$
v^2=2as\\
v=\sqrt{2as}
$$
Where s=2.2.
This is just another way to check.

Good Luck!