What is the momentum of the box at the bottom?

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The discussion revolves around calculating the momentum of a box sliding down a 35° incline, starting from rest. The box, weighing 98.0 N, encounters a frictional force of 32.0 N over a distance of 25.0 m. Initial calculations neglected the energy lost to friction, leading to an incorrect momentum value of 168 kgm/s. The correct approach involves accounting for the work done against friction in the energy balance equation. Ultimately, the accurate momentum at the bottom of the incline is confirmed to be 1.1 * 10^2 kgm/s.
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A 98.0 N box slides 25.0 m along a 35.0° incline. If the force of friction along the incline is 32.0 N, and the box starts from rest at teh top, what is the momentum of the box at the bottom?

This is how i attempted to solve this question. :confused: The answer in the book is 1.1 * 10^2 Kgm/s. Can you please tell me waht error i have made. Thank-you.

Eptop + Ektop = Epbot + Ekbot
mgh + 0 = 1/2mv^2 + 0
(9.81m/s^2)(25.0m*Sin35) = 1/2v^2
v=16.77 m/s

Fg=mg
m=98.0 N/9.81m/s^2
m=9.9898 Kg

p=mv
p=(9.9898 Kg)(16.77 m/s)
p=168 Kgm/s
 
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You didn't take into account the energy lost to friction.
 
Lol shaani nu gaal kadh :wink:
 
i think i figured it out:

Ep=Ek + Ffd
mgh=1/2mv^s + Ffd

THen solve for v and use that in p=mv
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

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