W=Fd, Work & Newton's Second Law problem

In summary: I got the right answer!In summary, the problem involves finding the force and work needed to lift a helicopter with a given acceleration and distance, using the equations W=Fd and F=ma. By considering the forces acting on the helicopter, the solution can be found by solving for the applied force, F_applied, which is equal to the mass of the helicopter times the acceleration due to gravity plus the mass of the helicopter times the given acceleration.
  • #1
Senjai
104
0
[SOLVED] W=Fd, Work & Newton's Second Law problem

Homework Statement



a) Find the force to give a helicopter of mass M an acceleration of 0.10g upward
b) Find the work done by this force as the helicopter moves a distance of h upward.

Homework Equations



[tex]W=F_{net}d[/tex]
[tex]F=ma[/tex]

The Attempt at a Solution



Personal Note: Q10 p. 145 Challenger.

This question isn't in the answer key, this post is mostly to verify my answer and figure it out if it's wrong.

I first stated F = Ma, then stated a = 0.10 x 9.8 m/s^2 = 0.98 m/2^2

Then for my answer, i had [tex]F=M(0.98_{m/s^2})[/tex]

would that be the proper way to state that answer?

b)
I stated W = Fd
Then stated W = M(0.98m/s^2)h

is this correct?

Thanks,
Senjai
 
Last edited:
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  • #2
For part a), the net force=0.1g

Remember:

[tex]0.1g=F_{net}=\sum F[/tex]

You should be able to find the correct, applied force on the helicopter.

Part b) is done correctly, except you need to use the correct [itex]F_{applied}[/itex]
 
  • #3
Oh, this isn't the first time questions worded with g and a coefficient has stumped me.

g = 9.8 m/s^2 right?

oh.. so your saying an acceleration of 0.1g so F = 0.1(9.8)?

I have trouble understanding exactly how it's asking for what... it's asking.
 
  • #4
Senjai said:
Oh, this isn't the first time questions worded with g and a coefficient has stumped me.

g = 9.8 m/s^2 right?

oh.. so your saying an acceleration of 0.1g so F = 0.1(9.8)?

I have trouble understanding exactly how it's asking for what... it's asking.

You aren't asked about the net force. You are asked what force it would take to raise the helicopter with a constant acceleration of [tex]0.1 g[/tex]

What forces act on the helicopter? There's [tex]F_{applied}[/tex] as GO1 mentioned, but there is another force acting on the helicopter as well.

Find out what force that is, plug everything into Newton's Second Law and isolate for [tex]F_{applied}[/tex] from there on out, you're good to go. :)
 
  • #5
So: Fapplied = M(0.1g) + mg?
 
  • #6
Senjai said:
So: Fapplied = M(0.1g) + mg?

Correct!, except the "m" in the second term, should be "M" in the notation of the problem. There is only one mass, M.
 
  • #7
Thanks for all your help guys!
 

Related to W=Fd, Work & Newton's Second Law problem

What is the formula for work (W=Fd)?

The formula for work is W=Fd, where W represents work, F represents force, and d represents displacement.

What is the unit of measurement for work?

The unit of measurement for work is joules (J).

What is Newton's Second Law of Motion?

Newton's Second Law of Motion states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass.

How is Newton's Second Law used to solve problems?

To solve problems using Newton's Second Law, you must first identify the net force acting on the object and its mass. Then, you can use the formula F=ma to calculate the acceleration of the object. Finally, you can use this acceleration to solve for other quantities, such as velocity or displacement.

Can work be negative?

Yes, work can be negative. This occurs when the force and displacement are in opposite directions, resulting in a negative value for work. Negative work can also represent work being done against the motion of an object.

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