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Finding Weight- Newton's Laws

  • Thread starter ledhead86
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  • #1
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A spacecraft descends vertically near the surface of Planet X. An upward thrust of 25.0 kN from its engines slows it down at a rate of 1.2 m/s^2, but if an upward thrust of only 10.0 kN is applied, it speeds up at a rate of 0.80 m/s^2.

Apply Newton's second law to each case, slowing down or speeding up, and use this to find the spacecraft's weight near the surface of Planet X.

I am completely confused. Any help would be appreciated.
 

Answers and Replies

  • #2
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I know that f=m*a, and w=m*g, but other than that im completely lost.
 
  • #3
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That's really oddly worded.

I think it's falling with a certain speed and is being pulled on by the planet's gravity. If it has a thrust of 25 kN then it'll actually accelerate away from the planet at -1.2 m/s^2, where the negative sign denotes going away from the planet. However, if it has a thrust of 10 kN then the ship will still be pulled towards the planet at: [Gravity - Acceleration due to thrust = 0.8 m/s^2]

Write an equation for the 25 kN force like the one above. Looks like you can use those equations and [(change in) F = M * (change in) A]
 
  • #4
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Skippy said:
That's really oddly worded.
I think it's falling with a certain speed and is being pulled on by the planet's gravity. If it has a thrust of 25 kN then it'll actually accelerate away from the planet at -1.2 m/s^2, where the negative sign denotes going away from the planet. However, if it has a thrust of 10 kN then the ship will still be pulled towards the planet at: [Gravity - Acceleration due to thrust = 0.8 m/s^2]
Write an equation for the 25 kN force like the one above. Looks like you can use those equations and [(change in) F = M * (change in) A]
you must be reading the problem wrong. It is descending at a certain rate. If they fire their boosters with a thrust of 25 kn then they will slow their descent to a rate of 1.2 m/s^2. If they only fire there boosters with a 10 kn thrust, their descent will continue to have an acceleration of .8 m/s^2.
They are moving towards the surface in both situations. But in the first situation, their velocity is slowing, and in the second situation, their velocity is getting faster.
 
  • #5
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bump......
 
  • #6
Tom Mattson
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Sum the forces on the ship. That shouldn't be too hard because there's only two: the thrust and gravity.

I know that f=m*a,
That's a little too simple. It's the sum of the forces that equals ma. Find the sum of the forces, set it equal to ma, and solve for m. Once you know m, finding mg should be easy.
 
  • #7
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Are you saying I should add 25+10, and 1.2=.8. And then plug those into f=ma to find m, which would be 17.5? Still confused.
 
  • #8
59
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Also, the direction of the acceleration in the case of 25kn thrust would be upward and the direction of the acceleration in the case of the 10kn thrust would be downward? Correct?
 
  • #9
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I think newton's secondly law says that forces on an object are its mass by acceleration ... so..
a being grav for planet X
thrust 1 + weight of obj to planet X is the mass of the object times its accel ...
F_g = m * accel due to planet X's gravity
25.0 kN + F_g = m * 1.2 m/s^2
and, similarly but with thrust 2
10.0 kN + F_g = m * -0.80 m/s^2
Some people might get antsy with the units, but I think i'm consistent enough to where as long as we have a positive mass it'll all be alright
.. 2 variables, 2 equations

this problem isn't really... true to its roots because gravity is different at different altitudes.. and being outside of planet X means that the g value isn't the actual gravity... so the question kind of takes a bigger picture and crams it into grain of sand

at least, thats how i'd do it.
 
  • #10
Use the difference in accelerations, and the difference in forces, to find the mass.
 
  • #11
I ended up with
w=16000N
g=~2.1

(highlight to view)
 
  • #12
142
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well... you know that the force of the thrust- force of the weight = ma...

So:

Fthrust-Fweight=m*a

work with that.
 

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