Momentum/Impulse homework question

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A 46.5 kg kid on roller blades pushes against a wall, resulting in a push duration of 0.460 seconds and a subsequent distance of 14.3 m before coming to rest due to friction, with a coefficient of friction of 0.125. Initial calculations for force and acceleration were misapplied, as they pertained to the rolling phase rather than the push against the wall. The impulse, defined as the change in momentum, should reflect the velocity gained after the push, not zero. To find the force exerted on the wall, the impulse can be calculated using the velocity obtained and the push duration. The correct approach involves determining the change in momentum and solving for the push force accordingly.
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Homework Statement



A 46.5 kg kid on roller blades pushes as hard as he can on a wall which causes him to start rolling away from the wall. At this point, friction begins to slow him down and he comes to rest 14.3 m from the wall. The coefficient of friction is 0.125. If his push lasted 0.460 s, how hard did he push on the wall (in Newtons)?

Rough diagram:
PAxhj.png


2. The attempt at a solution

J = Δ p
fΔt = mv2 - mv1 (mv2 = 0, since v2 is 0)
v1 = (Fft)/(-46.5)
v1 = -0.563 m/s **

Ff = μFn
=(0.125)(46.5)(9.8)
= 56.96N ( **I substituted this value to Ff in the (Fft)/(-46.5) equation to find v1)

a = v2 - v1 (v2 = 0)
t​
a =0.563
0.460​
a= 1.225 m/s2

Fnet = ma
= (46.5)(1.225)
= 56.96N
Therefore, the average force exerted on the wall is 56.96N(BKWD)?

I tried doing this problem today but I feel that the answer is wrong(doesn't make sense) and that I made a mistake. Any help would greatly be appreciated :-p
 
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It looks like all your calculations are for the period during which the kid is rolling. I.e., the acceleration you found is due to friction, not the force of the wall (in fact, at this point he's already let go of the wall)
 
JaWiB said:
It looks like all your calculations are for the period during which the kid is rolling. I.e., the acceleration you found is due to friction, not the force of the wall (in fact, at this point he's already let go of the wall)

So what do I need to change/add?

I understand what you are telling me, but I don't know what do I do
 
His initial momentum is zero (I guess), when he's done pushing on the wall, he has the velocity that you calculated. So what's the impulse?
 
JaWiB said:
His initial momentum is zero (I guess), when he's done pushing on the wall, he has the velocity that you calculated. So what's the impulse?

Impulse = Δ p
and since Δ p = 0, then J = Ff x t which is 56.96 x 0.460 which gives you 26.20 N.s

What do I do with the impulse value that I got?
 
Last edited:
Bump, can anyone help me out here?
 
Why did you say the change in momentum is zero? Before he pushes off the wall, he has no speed, and after he has some nonzero speed and therefore nonzero momentum. And if the change in momentum were zero, then J = 0.

J = F_push * t_push

You have t_push, you can calculate J from the change in momentum (which you can find using the velocity you calculated) so then you just have to solve for F_push
 

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