Questions on Conservation of Enery

[xxnicky]

first question:

Homework Statement

A 5.5e4 kg space probe is traveling at a speed of 11000 m/s through deep space. Retrorockets are fired along the line of motion to reduce the probe's speed. The retrorockets generate a force of 4.5e5 N over a distance of 2500 km. What is the final speed of the probe?

Homework Equations

W=FDcos(theta)
KE=1/2mv^2
PE=mgh

The Attempt at a Solution

(4.5e5N)(2500000m)cos0=1.125e12
(5.5e4)(9.8)(11000)=5929000000
1.125e12-5929000000=1.11907e12

I don't have any clue how to do this problem... I've attempted many times and got several different answers that are all wrong including 6386269.138, 6386269138, 9824.33, 4947.530329, 6396.021491, 4603.978509, and 6379.144998.



second question:

Homework Statement

A 47.0 g golf ball is driven from the tee with an initial speed of 50.0 m/s and rises to a height of 23.6 m.
What is its speed when it is 6.0 m below its highest point?

Homework Equations

W=FDcos(theta)
KE=1/2mv^2
PE=mgh

The Attempt at a Solution

KE=47.87984
47.87984-6=41.87984
I figured there would need to be more work involved than just that but I don't even know where to begin.


third question:

Homework Statement

A 54.5 kg. skateboarder starts out with a speed of 1.70 m/s. He does +80.0 J of work on himself by pushing with his feet against the ground. In addition, friction does -265 J of work on him. In both cases, the forces doing the work are nonconservative. The final speed of the skateboarder is 6.10 m/s.
(a) Calculate the change (changePE = PEf - PE0) in the gravitational potential energy.
(b) How much has the vertical height of the skater changed?

Homework Equations

W=FDcos(theta)
KE=1/2mv^2
PE=mgh

The Attempt at a Solution

-265-80=-345
I tried to attempt b and got 11295.9455 ans 2 as my answers but I don't remember exactly what i did to get them.


Sorry about posting 3 questions! I know that's a lot for one post. I have this online homework assignment due by 11:59 pm and I currently have a 17.84/20. I'm very frusterated with these three problems and no one in my class can get them either. I was hoping someone here could help. Thank you all so much! Your help is greatly appreciated.

 

[kamerling]

Use conservation of energy. We are in deep space far from any attracting masses, so the potential energy is 0. The equations you need are KE_intial + Work = KE_final
and W = FDcos(theta). theta is the angle between the velocity and the force vector. if these are pointing in the same direction theta = 0 and cos(theta) = 1. If they are pointing
in the opposite direction ...

 

[xxnicky]

I'm not sure if i understand exactly what you're saying... like I'm not sure how to relate what you said to the problems. Will you please explain it again?
thank you for posting! sorry about this!

 

[kamerling]

you can use conservation of energy. The potential energy is 0, so there's just kinetic
energy. You have an initial kinetic energy which you can compute with (1/2)mv^2.
Now you have a force which does work on the spacecraft . If this work is positive, the kinetic energy will increase, if this work is negative it will decrease. The formula for work is
FDcos(theta) F is the force, D is the distance over which the force acts, theta is the angle between the force and the velocity of the spacecraft .

 

[xxnicky]

thank you so much!