please help me with this problem. i know it's VERY easy, but i can not find the right answer. please help.

Question

A cart with mass 19.5 kg is intitally at rest. You get it moving by pushing on the cart at an angle of θ = 33.9° from above the horizontal. The magnitude of your force as a function of time is given by: Fyc(t) = Foe-b t, where b = 0.46 s-1 and Fo = 141.6 N. You can assume that the wheels on the cart are good enough that you can neglect friction between the cart and ground.
What is the speed of the cart when t = 4.7 s?

I know that i need to use newton's second law. I worked on it for hours, but can not get it. please help. Thank you so much.

i entered 12.6 m/s into the system and it says it's wrong. I can not use the concept of impulse here as i did not study it yet. can you solve it using newton's second law??
Thanks anyway...

Answer 1

The key to this question is to use the concept of impulse.

Horizontal force = applied force × cos 33.9°.
Total horizontal impulse
= ∫F(t) dt
= ∫(t = 0 to 4.7) 141.6 (cos 33.9°) e^(-0.46 t) dt
= 117.5 [e^(-0.46 t) / (-0.46)][0 to 7]
= -255.5 [e^(-3.22) - 1]
= 245.3 Ns

Impulse = Δp, so the final momentum is 245.3 Ns = 245.3 kg m/s = mv = 19.5 v. So v = 245.3/19.5 = 12.6 m/s.

[ADDITIONAL]
You can get around it without using impulse, but it's a little messier.
Horizontal force = applied force × cos 33.9° as before, so we have
F_h = 141.6 (e^-0.46t) cos 33.9°
= 117.5 e^(-0.46t)
By F = ma, we get a = 117.5 e^(-0.46t) / 19.5
= 6.027 e^(-0.46t).

v(4.7) - v(0) = ∫(0 to 4.7) a(t) dt
so v(4.7) = ∫(0 to 4.7) 6.027 e^(-0.46t) dt
= [6.027 e^(-0.46t) / (-0.46)] [0 to 4.7]
= -13.10 (e^(-2.162) - e^0)
= -13.10 (-0.8849)
= 11.6 m/s.

I accidentally changed 4.7 to 7 part way through my first answer - the last few lines should be as follows:
= 117.5 [e^(-0.46 t) / (-0.46)][0 to 4.7]
= -255.5 [e^(-2.162) - 1]
= 226.1 Ns
and then v = 226.1/19.5 = 11.6 m/s.