How Much Force Does Pam's Rocket Pack Generate?

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Pam's rocket pack generates a force of approximately 3.5 x 10^3 N, calculated using kinematic equations and Newton's second law. Initially, a common mistake was to confuse this force with her weight, which is 392 N. The final kinetic energy of Pam after using the rocket pack is calculated to be 7.69 x 10^4 J. The discussion highlights the importance of correctly applying physics principles to solve for acceleration and force. Overall, the calculations demonstrate the relationship between force, mass, and acceleration in a frictionless environment.
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Homework Statement



Pam has a mass of 40.0kg and she is at rest on a smooth level frictionless ice. pam straps on a rocket pack. the rocket pack supplies a constant force for 22.0m and pam acquires a speed of 62.0m/s

a)what is the magnitude of the force?
b)what is pams final kinetic energy

Homework Equations



a)F=mg
b)Ek=1/2mv^2



The Attempt at a Solution



a)
m=40.0kg
d=22.0m
v=62.0m/s
F=mg
F=(40.0)(9.8)
F=392N

b)
m=40.0
v=62.0
Ek=?
Ek=1/2mv^2
Ek=1/2(40.0)(62.0^2)
Ek=7.69x10^4J
 
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Iceclover said:

Homework Statement



Pam has a mass of 40.0kg and she is at rest on a smooth level frictionless ice. pam straps on a rocket pack. the rocket pack supplies a constant force for 22.0m and pam acquires a speed of 62.0m/s

a)what is the magnitude of the force?
b)what is pams final kinetic energy

Homework Equations



a)F=mg
b)Ek=1/2mv^2



The Attempt at a Solution



a)
m=40.0kg
d=22.0m
v=62.0m/s
F=mg
F=(40.0)(9.8)
F=392N

This is not the force you were asked for: this is the weight of the person. Try using the relevant kinematic equation to calculate the acceleration, then use Newtons 2nd law to work out the force.
b)
m=40.0
v=62.0
Ek=?
Ek=1/2mv^2
Ek=1/2(40.0)(62.0^2)
Ek=7.69x10^4J

This looks fine
 
how do i figure out the acceleration without the force?
 
wow i definitely did not do this right, i got 3480
 
Iceclover said:
how do i figure out the acceleration without the force?

Using the kinematic equations [hint: the relevant one to use here is v2=u2+2as, where here v is the final velocity, u is the initial velocity, a is acceleration and s is the distance.]
 
Iceclover said:
wow i definitely did not do this right, i got 3480

I get 3495N
 
i got 3520
 
ok so if you put it into sig figs you would get 3.5x10^3
 
Iceclover said:
ok so if you put it into sig figs you would get 3.5x10^3

Yes, if you're using 2sf.
 
  • #10
ok so that's right then?
 
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