Force in kinetic energy/force in momentum =1

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SUMMARY

The discussion centers on the equivalence of force calculated from kinetic energy and momentum using a mass of 20 lb, an initial velocity of 0 ft/sec, a final velocity of 10 ft/sec, and a time interval of 2 seconds. The kinetic energy (K.E.) is calculated as 1000 lb(ft/sec)2, leading to a force of 100 lb when divided by the distance of 10 ft. Similarly, momentum is calculated as 200 lb(ft/sec), resulting in the same force of 100 lb when divided by the time of 2 seconds. The calculations confirm that the force derived from both kinetic energy and momentum is indeed equal.

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  • Understanding of basic physics concepts such as kinetic energy and momentum
  • Familiarity with the equations K.E. = 1/2 mv2 and momentum = mv
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Homework Statement


mass=20
V1=0
V2=10ft/sec
t=2 sec
a=5'/sec/sec
x=10ft
I want to show that the force in momentum and in kinetic energy from above data is the same. With the understanding that simple algebra is being superimposed on calculus.


Homework Equations


K.E.= integral of force over distance= max=mdv/dt (dx) =1/2mv^2
momentum = integral of force over time = mv
force= K.E./change in distance
force=momentum/change in time



The Attempt at a Solution


K.E.= 1000 lb (ft/sec)^2
momentum=200 lb(ft/sec)
The force from kinetic energy = 1000 lb (ft/sec)^2 /10ft =100 lb
The force from momentum = 200 lb (ft/sec)/2 sec= 100 lb
Same force, correct ?


 
Last edited:
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I am not sure how your units work, but lbs is a force, so you'd need to divide by acceleration due to gravity to get the mass.
 
rock.freak667 said:
I am not sure how your units work, but lbs is a force, so you'd need to divide by acceleration due to gravity to get the mass.

OK the mass =20 by correction.
Now is the 100 lb force correct ?
 

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