How to calculate distance when there's a changing force?D:

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To calculate the distance a box moves under a changing force, one must apply calculus to derive the acceleration from the force function F(t) = (6.00 N/s^2)t^2. The mass of the box is 3.50 kg, leading to an acceleration a(t) = (6.00 / 3.50)t^2. To find the velocity and position functions, integrate the acceleration over time, starting from the initial conditions. The initial speed is 10.0 m/s, and the calculations must account for the force's effect until the box's speed reaches zero. Proper application of these principles will yield the correct distance and final speed after 4 seconds.
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How to calculate distance when there's a changing force?D:

Homework Statement



A 3.50kg box is moving to the right with speed 10.0m/s on a horizontal, frictionless surface. At t = 0 a horizontal force is applied to the box. The force is directed to the left and has magnitude F(t) = (6.00N/s^2)t^2

What distance does the box move from its position at before its speed is reduced to zero?

If the force continues to be applied, what is the speed of the box at 4.00 ?

Homework Equations



x = x0 + v0xt + .5axt^2
F = ma
vx = v0 + axt
I might be missing several

The Attempt at a Solution



I don't even know where to start for this problem.
I drew a free body diagram and it didn't help me.
What I calculated :
Force downward is 3.5kg x 9.8 m/s^2
or the weight
I don't know how to calculate the forces to the left and right.

For the second question I plugged in 4 to the F(t) equation and got 96
so I used 96 = ma SO
96 = 3.5 * a
a = 27.4
and I plugged 27.4 into vx = v0x + axt
so vx = 10m/s + 27.4(4)
and got 119.7 and this is not correct.

Help please, thank you!
 
Last edited:
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I tried googling the problem and someone else had a similar problem. They got 6.00 for acceleration and I'm not too sure as to how the person got it, but I used it and got 19.5 for the first question.

I don't know how to calculate the new velocity after 4 seconds though.
I used

vx = v0 + axt
vx = 10m/s + 6.00(4)
vx = 34...incorrect.
please help
 


You'll need to use calculus. You start here:

F(t) = (6.00N/s^2)t^2 = (6.00 kg m / s^2) t^2
m = 3.50kg
a(t) = (6.00 / 3.50) (m/s^2) t^2
v(t) = ?
x(t) = ?
 
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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