sorry! I never noticed..the mass should be 52000kg sorry about that!
For Further Reading
physics repost - Alexis, Sunday, October 14, 2007 at 6:07pm
so it should be a 52000kg train slows from 25 m/s to 14m/s in 5.0 s. Calculate the work done on the train.
so I found the displacement by using
displacement= 1/2 (v1+v2) (delta t)
and got 97.5
and then I found the force which is F=mg
and got 510120N, then took that force and did W=Fd
and got 49736700J
my textbook says 1.1x10^7J? How is that right? I think I am right?
physics repost - Alexis, Sunday, October 14, 2007 at 6:17pm
I tried to get the answer using the kinetic energy formula, and it works (for the answer of 1.1x10^7), but the only problem is, the homework is for the pages before the kinetic energy, so I would like to know how to figure out the answer without the kinetic energy formula. Thank you for helping me!
HAHA I just got the answer!!!! You don't use F=mg it should be ma!!! Thanks anyways!!
To calculate the work done on the train without using the kinetic energy formula, you need to find the net force acting on the train and multiply it by the displacement.
First, let's calculate the net force on the train.
Given:
Mass of the train (m) = 52000 kg
Initial velocity (v1) = 25 m/s
Final velocity (v2) = 14 m/s
Time taken (delta t) = 5.0 s
To find the net force, you can use Newton's second law of motion:
F = ma
To find the acceleration (a), use the equation:
a = (v2 - v1) / (delta t)
a = (14 m/s - 25 m/s) / 5.0 s
a = -11 m/s / 5.0 s
a = -2.2 m/s^2
Now substitute the mass (52000 kg) and the acceleration (-2.2 m/s^2) into the equation F = ma:
F = (52000 kg) * (-2.2 m/s^2)
F = -114,400 N
Since the net force acting on the train is negative, it means there is a force acting against the direction of motion. This is because the train is slowing down.
Now, let's calculate the work done on the train.
We already have the net force (-114,400 N), and we need to find the displacement.
You calculated the displacement using the formula:
displacement = 1/2 (v1 + v2) (delta t)
displacement = 1/2 (25 m/s + 14 m/s) (5.0 s)
displacement = (39 m/s) * (5.0 s)
displacement = 195 m
Now, use the formula: work (W) = force (F) * displacement
W = (-114,400 N) * (195 m)
W ≈ -22,308,000 J
The negative sign indicates that work is done against the direction of motion, which makes sense since the train is slowing down.
Therefore, the work done on the train is approximately -22,308,000 J.
It's worth noting that the answer in your textbook (1.1x10^7 J) seems to be incorrect based on the given information and calculations above.