Excellent work!
Your calculations are correct, but near the end, you have forgotten to cancel u. If you cancel the u, the linear term becomes a constant, and the resultant equation becomes a simple quadratic which is much easier to solve.
I get u=7.468 m/s using g=9.81 (as you did)
You will rework and should get 7.468 m/s as well.
Congrats for the good work!
We have that the block is moving horizontally. Hence, its potential energy due to gravity stays the same. The only change in its mechanical energy is the one due to the change of speed. This reduction of its kinetic energy, due to the conservation of energy, is equal to the work that friction does. We have that at A the kinetic energy is : K=1/2*m*u^2=10*10*10/2=500J. At B, we have that K=1/2*10*16=80J. Sine we have that the initial value is 500, the work from the friction force (opposite to the movement of the object) is 80-500=420J.
Between the two asymptotic gigantic branches, the Sun changes the greatest in size, brightness, and temperature.
Answer:
Answered
Explanation:
a) What is the work done on the oven by the force F?
W = F * x
W = 120 N * (14.0 cos(37))
<<<< (x component)
W = 1341.71
b) 
= 29.4 N
W_f= 328.72 J = 329 J
c) increase in the internal energy
U_2 = mgh
= 12*9.81*14sin(37)
= 991 J
d) the increase in oven's kinetic energy
U_1 + K_1 + W_other = U_2 + K_2
0 + 0 + (W_F - W_f ) = U_2 + K_2
1341.71 J - 329 J - 991 J = K_2
K_2 = 21.71 J
e) F - F_f = ma
(120N - 29.4N ) / 12.0kg = a
a = 7.55m/s^2
vf^2 = v0^2 + 2ax
vf^2 = 2(7.55m/s)(14.0m)
V_f = 14.5396m/s
K = 1/2(mv^2)
K = 1/2(12.0kg)(14.5396m/s)
K = 87.238J
Balanced equation:
6 HCl + Fe2O3 = 2 FeCl3 + 3 H2O
