Net-Work Congestion - An Energy shortage?

7. Any time you are asked to find the minimum work, assume the work is done at equilibrium. Thus the force required to push the cart up the incline in (a) is mg sin(theta). The work is therefore mg*sin(17.5)*300m*(1) = 884,000 J

b) Again, using sigma_Fx = 0, the force required to do this is (mg*sin(17.5) + mu*mg*cos(17.5))* 300 m = 1,600,000 J.

13. The work done by the force is the area under the curve for the given start and end positions.
a) Work = .5*(400 N)*(10 m + 4 m) = 2,800 J
b) Work done from x = 10 to 15 m is NEGATIVE (since it is in the negative direction), so the total work will be 2,800 J - .5*(200 N)*(5 m + 2 m) = 2,100 J

18.
a) If the KE is doubled, the speed is multiplied by the square root of 2.
b) if the speed is doubled, the KE is multiplied by 4.

22. v = (2*Fnet*delta_x/m)^(1/2) = 43.59 m/s

23. Fnet = mv^2/(2*delta_x) = 343 N

25. Through a free body diagram, you should obtain that Fnet = mu*mg. Using this and the work-energy theorem:
mu*mg*delta_x*(-1) = 0 - .5*m*v₀^2
Solving:
v₀ = (2*mu*g*delta_x)^(1/2) = 26.91 m/s

28.
a) This is a Newton's 2nd problem! T = mg + ma = 2479 N
b) Wnet = Fnet*delta_x*(1) = m*a*delta_x = 6791 J
c) W_T = T*delta_x*(1) = 52,059 J
d) W_g = mg*delta_x*(-1) = -45,276 J
e) Wnet = .5*m*v_f² - 0

So v_f = (2*Wnet/m)^(1/2) = 7.852 m/s