Work

Jul 2008
179
1
Consider the two situations, where two blocks of equal masses down a runway without friction, since the height of hours until the soil. The blocks come with the same speed at the base? The work done by force weight on them is the same? The time duration of the movement is the same? Justify the answers.
 

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topsquark

Forum Staff
Apr 2008
3,115
661
On the dance floor, baby!
Consider the two situations, where two blocks of equal masses down a runway without friction, since the height of hours until the soil. The blocks come with the same speed at the base? The work done by force weight on them is the same? The time duration of the movement is the same? Justify the answers.
If there is no friction then according to the work energy theorem:
\(\displaystyle W_{nc} = \Delta E = 0 \implies KE_f = PE_i = mgh\)
where h is the initial height of the blocks over the final height. So the speed is the same for both blocks and the work done by gravity is the same for both. (I'll let you show that.)

As far as the time duration we have two variables: the distance traveled and the angle the block is moving at. We cannot say specifically, by looking at the diagrams, which block will get to the bottom faster, all we can say is that they get there at the same speed.

-Dan
 
Jul 2008
179
1
\(\displaystyle W1=m1gh\)

\(\displaystyle W2=m2gh\)

\(\displaystyle m1=m2\)
\(\displaystyle w1=w2\)

\(\displaystyle W1=\frac{m1.(V1)^2}{2}-0\)

\(\displaystyle W2=\frac{m2.(V2)^2}{2}-0\)

\(\displaystyle W1=W2; m1=m2\)
\(\displaystyle V1=V2\)

The time duration of the movement is the same? As I show?
 

topsquark

Forum Staff
Apr 2008
3,115
661
On the dance floor, baby!
\(\displaystyle W1=m1gh\)

\(\displaystyle W2=m2gh\)

\(\displaystyle m1=m2\)
\(\displaystyle w1=w2\)

\(\displaystyle W1=\frac{m1.(V1)^2}{2}-0\)

\(\displaystyle W2=\frac{m2.(V2)^2}{2}-0\)

\(\displaystyle W1=W2; m1=m2\)
\(\displaystyle V1=V2\)

The time duration of the movement is the same? As I show?
The speeds are the same, yes, and you did that correctly. The time it takes for each object to get to the bottom can only be figured out when the curve in the second graph is given. Otherwise we can make an educated guess, but it is impossible to tell.

-Dan
 

topsquark

Forum Staff
Apr 2008
3,115
661
On the dance floor, baby!
I think the time is different. It has some way to prove?
As I've been saying we need to know the exact characteristics of the curve in the second graph. Without that we cannot say what the time in the second graph will be.

For example, if one of the masses was sliding down a gentle slope right up until the final point, where it drops steeply, the time taken will be much greater than the straight slope as in the first graph.

-Dan