The potential energy of the box at the top of the incline is given by:
PE = mgh
where m is the mass of the box, g is the acceleration due to gravity, and h is the height of the incline. Since the box starts from rest, all of the potential energy is converted to kinetic energy at the bottom of the incline. The kinetic energy of the box is given by:
KE = (1/2)mv^2
where v is the velocity of the box at the bottom of the incline.
The height of the incline is given by h = Lsinθ, where L is the length of the incline and θ is the angle of incline. Substituting the given values, we have:
h = 3sin30° = 1.5 meters
The potential energy of the box at the top of the incline is:
PE = mgh = (10 kg)(9.8 m/s^2)(1.5 m) = 147 J
The work done by friction is given by:
W = Fd = (15 N)(3 m) = 45 J
The net work done on the box is:
W_net = PE - W = 147 J - 45 J = 102 J
This work is equal to the change in kinetic energy of the box:
W_net = ΔKE = KE_f - KE_i
Since the box starts from rest, the initial kinetic energy is zero. Therefore:
KE_f = W_net = 102 J
Substituting the mass of the box, we have:
KE_f = (1/2)mv^2
102 J = (1/2)(10 kg)v^2
Solving for v, we get:
v = sqrt(20.4) m/s
Therefore, the kinetic energy of the box at the bottom of the incline is 102 J, and the velocity of the box at the bottom of the incline is approximately 4.51 m/s.