Solving for Distance of Crate on Level Floor with Given Force & Time

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To determine the distance a 29.8kg crate moves when pushed with a force of 188.535 N on a level floor, the relevant physics principles include Newton's second law and the effects of friction. The crate's weight is calculated as 29.8kg multiplied by the acceleration due to gravity (9.8 m/s²), which helps establish the normal force. The frictional force is derived from the coefficient of kinetic friction (0.338) and the normal force. The net force acting on the crate is then found by subtracting the frictional force from the applied force, allowing for the calculation of acceleration. Finally, using the formula for distance, the correct distance can be calculated, ensuring all forces and accelerations are accurately accounted for.
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A 29.8kg crate is at rest on a level floor, and the coefficient of kinetic friction is 0.338. The acceleration of gravity is 9.8 m/s. if the crate is pushed horizontally with a force of 188.535 N, how far does it move in 5.88 seconds?

i really need help with this...thanks!
 
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So use Newton2 law's formulae: F=ma; s=(at^2)/2
 
so i did 188.535= (29.8) (A)
A= 6.33

D=(6.33)(5.88)^2 /2
=109.43

but that's not the right answer i don't know what I am doing wrong
 
The weight of the box is 29.8*9.81= (*)N
therefore the normal reaction,R= (*) N

the frictional force=uR where u=coefficient of friction.

find the resultant force and then equate that to ma
 
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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