Will the Box Move on a Braking Truck? Mathematical Justification Required

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AI Thread Summary
A 5 kg box on a pickup truck moving at 90 km/h is analyzed to determine if it will slide when the truck brakes uniformly to a stop over 125 meters. The maximum frictional force between the box and the truck is 15N. Calculations indicate the truck's acceleration is approximately -0.1 m/s², which is less than the acceleration due to gravity, suggesting the box will not move. The discussion emphasizes the importance of consistent unit usage and correctly applying the equations of motion. The conclusion is that the box remains stationary due to insufficient force overcoming friction.
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Homework Statement



A box with a mass of 5 kg sits in the middle of the bed of a pickup truck moving with a speed of 90 km/h. The maximum force of friction between the box and the truck is 15N. If the truck were braked uniformly to a stop in 125 m, would the box move? Justify your answer mathematically.

Homework Equations



vf2= vo2 + 2ax?

The Attempt at a Solution



a= (vf2 - vo2) / 2x
a= (0-9) / 2(125)
a= -0.36
yes is would
am i even close?
 
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Almost correct .

Calculate the acceleration of the truck - make sure you use consistent units.
Then draw a diagram of the forces acting horizontally on the box, remember f=ma
 
right so, 90 km/h is 25m/s. therefore acceleration [of truck] is equal to -0.1m/s2
 
Think you have a calculator error there
a = v2/ 2s = 252 / 2*125 ms-2
 
Thread 'Variable mass system : water sprayed into a moving container'

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

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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