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Last Minute Force Questions

  • Thread starter SkiingAlta
  • Start date
  • #1
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Hey all. Sorry for asking so many questions about force, but these are tricky to me! Okay so I have 2 questions that I have started, but have yet to complete. Help would be greatly appreciated.

1. A model rocket weighs 5.4 N.
a) What is its mass?
b) What applied force makes it accelerate upward at 13.0 m/s2?

Okay, so I did 5.4/9.8 to get mass which is 0.55 kg right?

How do I do b?


Then:
2. A space probe and rocket weigh 88,888 N. the applied force propelling the rocket is 3.0 x 106 N. Determine:
a) the mass of the rocket and space probe.
b) the upward acceleration of the rocket and space probe.
c) the velocity of the rocket 15 seconds after lift off.

So... I got a by going 88888/9.8 = 9070.20 kg

I don't know what to do now.
 

Answers and Replies

  • #2
598
0
Part a is correct.

For part b: You know the mass, which of Newton's Laws would you use?
 
  • #3
PhanthomJay
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Hey all. Sorry for asking so many questions about force, but these are tricky to me! Okay so I have 2 questions that I have started, but have yet to complete. Help would be greatly appreciated.

1. A model rocket weighs 5.4 N.
a) What is its mass?
b) What applied force makes it accelerate upward at 13.0 m/s2?

Okay, so I did 5.4/9.8 to get mass which is 0.55 kg right?

How do I do b?


Then:
2. A space probe and rocket weigh 88,888 N. the applied force propelling the rocket is 3.0 x 106 N. Determine:
a) the mass of the rocket and space probe.
b) the upward acceleration of the rocket and space probe.
c) the velocity of the rocket 15 seconds after lift off.

So... I got a by going 88888/9.8 = 9070.20 kg

I don't know what to do now.
Your calcs for the mass are correct. For the acceleration, use newton's 2nd law .
 
  • #4
19
0
Okay, so I use f=ma...

F=.55*13
F=7.15 N
That right?


=Then:
2. A space probe and rocket weigh 88,888 N. the applied force propelling the rocket is 3.0 x 106 N. Determine:
a) the mass of the rocket and space probe.
b) the upward acceleration of the rocket and space probe.
c) the velocity of the rocket 15 seconds after lift off.

So... I got a by going 88888/9.8 = 9070.20 kg

I don't know what to do now.
Now how should I start this?
 
  • #5
PhanthomJay
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Okay, so I use f=ma...

F=.55*13
F=7.15 N
That right?
Newton's law says that the sum of all forces , or the net force, equals mass times acceleration. So what you have calculated is the net force. The problem wants the upward applied force. There is more than one force acting.
Now how should I start this?
same way.
 
  • #6
598
0
Okay, so I use f=ma...

F=.55*13
F=7.15 N
That right?




Now how should I start this?
a is correct.
For part b, it is similar to part b in the first question. What do you know? Which equation would you use?
 
  • #7
19
0
So the other force acting would be the weight? Which would be 5.4 N. Subtract that from the upward force? I'm confused....
 
  • #8
2
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SkiingAlta,
Your calculation to find mass is correct, the answer is .55kg. For part b, the universal equation for Force is F=ma. In this question, you would multiply 13.0 m/s² (the acceleration) by .55kg (the mass) to get a Force of 7.15 N.

For the second question, your calculation to find the mass was again correct. For part b, you would use the same equation F=ma, but in this case, it would be for solving for acceleration. For this, you would do 3.0 x 10^6 N (the Force) divided by 9070.20 kg (the mass) to get your answer.
 
  • #9
PhanthomJay
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I'm afraid you are getting some incorrect responses.
So the other force acting would be the weight?
Yes.
Which would be 5.4 N.
yes again.
Subtract that from the upward force? I'm confused....
no. The upward force is the applied force; the downward force is the weight. Since the acceleration is upward, the net force must be upward, and the applied force up must be bigger than the down force, and algebraically, they must sum to 7.15 N up. So what is the applied force?
 
  • #11
PhanthomJay
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8895.15?
Where did that number come from??????
Let the applied force = P. Then ,
P - 5.4 = F_net = 7.15. Solve for P.
Now try part b.
 
  • #13
PhanthomJay
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12.55 n.
Yes, the applied force minus the weight force is the net force, so P = 12.55 N. Acceleration is always in the direction of the the net resultant force. Problem 2 works the same way, except that this time you are given both forces, and are asked to find the acceleration, using F_net = ma. Once you get the acceleration, you can find the speed after 15 sec using one of the the kinematic motion equations.
 

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