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Here’s my work to your question. I used Newton’s Second Law and a kinematics equation to arrive at the answer.
Answer:
Explanation:
I can tell you what the answers for the middle column are, but if you don't know how to solve total energy problems, they won't make any sense to you at all.
First row, KE = 0
Second row, KE = 220500 J
Third row, KE = 183750 J
Fourth row, KE = 205800 J
That's also not paying any attention to significant digits because your velocity only had 1 and that's not enough to do the problem justice. I left all the digits in the answer. Round how your teacher tells you to.
Answer:
c) F = 16000 N
Explanation:
For this exercise we use Newton's second law
F = ma
they tell us that adding the other wagons the acceleration of the locomotive must be maintained
F = m a
by adding the other four wagons
mass = 4 no
therefore to maintain the force you must also raise the same factor
Fe = 4Fo
Fe = 4 4000
F = 16000 N
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vo = 25 m/sec
<span>vf = 0 m/sec </span>
<span>Fμ = 7100 N (Force due to friction) </span>
<span>Fg = 14700 N </span>
<span>With the force due to gravity, you can find the mass of the car: </span>
<span>F = ma </span>
<span>14700 N = m (9.8 m/sec²) </span>
<span>m = 1500 kg </span>
<span>Now, we can use the equation again to find the deacceleration due to friction: </span>
<span>F = ma </span>
<span>7100 N = (1500 kg) a </span>
<span>a = 4.73333333333 m/sec² </span>
<span>And now, we can use a velocity formula to find the distance traveled: </span>
<span>vf² = vo² + 2a∆d </span>
<span>0 = (25 m/sec)² + 2 (-4.73333333333 m/sec²) ∆d </span>
<span>0 = 625 m²/sec² + (-9.466666666667 m/sec²) ∆d </span>
<span>-625 m²/sec² = (-9.466666666667 m/sec²) ∆d </span>
<span>∆d = 66.0211267605634 m </span>
<span>∆d = 66.02 m</span>