Posted earlier & still can't figure/don't have correct answer...can you help?

Question

Man pushes a 22 kg box across a frictionless floor. The box is initially at rest. He initially pushes on the box gently, but gradually increases his force so that the force he exerts on the box varies in time as F = (8.6 N/s)t. After 3 s, he stops pushing. The force is always exerted in the same direction.

What is the velocity of the box after 3 s? answer in m/s

How far has the man pushed the box in 3 s? answer in m

What is the average velocity of the box between 0 s and t? answer in m/s

What is the average force that the man exerts on the box while he is pushing it? answer in N

HELP!

Answer 1

Man is gonna have a helluva time pushing that box on a frictionless floor, so we'll have to ignore how he does it.

F = ma = kt, k=8.6N/s
a = dv/dt = kt/m
dv = ktdt/m
1)
v= (k/(2m))t^2+u
u = 0
v(3) = 8.6*9/(2*22) +0 m/s
v = ds/dt =(k/(2m))t^2+u
2)
s = (k/(6m))t^3 + ut + s0
s(3) = 8.6*27/(6*22) + 0 + 0 m
3)
average v = (integral(vdt))/t = s/t
average v = 8.6*9/(6*22) m/s
4)
average F = (integral(kdt))/t = (k/2)t
average F = 8.6*3/2 N

Answer 2

because the acceleration is not constant, you would normally use calculus to determine the force over time

velocity is the integral of acceleration
distance is the integral of velocity

F=ma so a=F/m

a=8.6/22 m/sec^2= .391 m/sec^2(t)

that is, you would integrate the acceleration function with respect to time giving (1/2)*.391t^2 (the constant is neglected because their was no initial force)

after 3 seconds, v=.1955(3)^2= 1.76 m/s

integrating .1955t^2 we get (1/3).1955(3)^3 (the constant again disappears because of zero distance start) = 1.76 m

isn't that cute how they set that up?

the average speed is the total distance divided by the total time

.0652t^3/.1955t^2 = .334 t

Now, assuming I have remembered any of my college physics correctly, and assuming I have done any of the calculus and algebra correctly, then you should be able to do the last one based on what is here.

If I have screwed it up (and that is not unlikely) then it won't help you for me to do the last one anyway.

remember that the average force can be obtained from the average acceleration (F=ma) and the average acceleration is equal to the final velocity divided by the total time

go for it

again, I am not very confient of this solution but it may put you on the right track if I remembered even part of it correctly

Answer 3

F = M(dv/dt) = 8.6 t; so that Mdv= 8.6 t dt

Then dv = K t dt; where K = 8.6/M

v final = K (t^2/2), by integrating over t = 0 through 3 sec and v = 0 through v final; so that v final = (8.6/22)(9/2) m/sec

dS/dt = v final = K (t^2/2) and so dS = K (t^2/2) dt giving S = (K/2)(t^3/3); so that S = (K/2)(9/3) meters after 3 sec.

v avg = S/t = [(K/2)(9/3)]/3 = K/2 m/sec

a avg = v final/3 sec; so that F avg = M(a avg) = M (v final/3) Newtons

You can do the math.

Answer 4

F = ma = m(dv/dt)

so, dv = 8.6t/22 dt

Integrate from 0 to v on left and 0 to 3 on right.

So, v = 8.6*9/44 m/s = 1.76 m/s

ma = m (d2x/dt2)

So, d2x = 8.6t/22 dt2

And integrate.

Answer 5

Use integration.

Its important to note that accelleration is not constant here.
F=ma,
m=22kg, F=(8.6N/s)t
a=F/m
a=(8.6N/22kg*s)t
integrate for velocity:
v=(4.3/22)t^2 + C. C is zero because initial velocity at t=0s was zero.
Integrate for position, x:
x=(4.3/66)t^3 + C. C is zero here because initial position was zero too.

Now you have all your equations you should be able to solve it.
For average velocity, you take your total distance traveled and divide by total time. For average Force you take maximum force at t=3, and divide it by 2.