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.
2007-09-09 21:57:38 · 3 answers · asked by moonlightNY 1 in Science & Mathematics ➔ Physics
i believe that i need to summ all the forces in the x direction, solve for acceleration, then integrat to find velocity. i am kinda lost here!!!!
2007-09-09 21:58:44 · update #1
Consider another approach: Use Work - Energy. Calculate the total work done on the cart in the form t = 0 to t = 4.7 seconds. Since we are neglecting friction all work will be transferred to kinetic energy. Use that to find the velocity.
2007-09-09 22:23:22 · answer #1 · answered by joe_ska 3 · 0⤊ 0⤋
F = F0e^(-bt) ...... at least I think this is what you meant
If A is the angle to the horizontal (= 33.9 degrees) then:
Fx = Fcos(A) = F0e^(-bt)cos(A) = ma = mdv/dt
dv/dt = (1/m)(F0e^(-bt)cos(A))
dv = (1/m)(F0e^(-bt) cos(A))dt
Integrate:
v(t) - v(0) = (1/m)(-bF0 e^(-bt)cos(A)) evaluated from t=0 to 4.7
v(0) = 0 since it start from rest
v(t) = (bF0cos(A)/m)[-e^(-4.7b) - (-e(0))]
v(t) = (bF0cos(A)/m)[1 - e^(-4.7b)]
v(t) = (2.77)[1 - e^(-2.162)] = 2.77[1 - 0.115]
v = 2.45 m/s
2007-09-09 22:31:26 · answer #2 · answered by Captain Mephisto 7 · 0⤊ 0⤋
i can't help you
2007-09-09 22:22:28 · answer #3 · answered by Anonymous · 0⤊ 0⤋