Question

Sam,whose mass is 80 kg takes off down a 60 m high,12 degree slope on his jet powered skies. The skies have a thrust of 180 newtons. The coefficient of kinetic friction of his skies on the snow is .0650

a) draw a free body diagram for  Sam( on his skies)

b) what is the magnitude of the force of friction on Sam 

c) what is the magnitude of Sam's acceleration

d) what is Sam's speed at the bottom of the slope

Answer 1

38 figures in geometric phisics biologically phisical mortal phisics.

Answer 2

how many

Answer 3

outragious

Answer 4

enargy protrators

Answer 5

compass one destination to another

Answer 6

hevier or lighter

Answer 7

Assuming that Sam's initial velocity is 0, his total energy is all gravitational potential energy = T = m g h.
(b) The normal force is m g cos(theta), and the frictional force F is (mu m g h), where m is the combined mass of Sam and his jet skis, g = 9.81 m/s^2,
h = 60m, mu = .0650, and theta = 12 degrees,
so F = (.0650)(80 kg)(9.81 m/s^2) = 49.897 N.

(c) The component of Sam's weight along the 12 degree incline is
m g sin(alpha) = (80)(9.81)(sin(12 degrees)) = 163.17 N = F1.
The component of Sam's weight and the thrust P act down the incline, and the frictional force acts up the incline, so the net force acting on Sam is
F1 + m g sin(alpha) - F = 293.27 N = m a
a = 293.27/80 = 3.6659 m/s^2.

(d) Once Sam is at the bottom of the incline, all his energy is kinetic. Thrust adds to Sam's initial total energy, and friction detracts from it. The amount of work W1 done by the frictional force on Sam = -F*s, where s is the length of the incline and is 60/sin(12 degrees) = 288.58 m, so W1 = -14399.56 J.
The amount of work W2 done by gravity on Sam is m g h = 47088 J. The work W2 done by thrust P is P*s = 51945 J. Once Sam reaches the bottom of the incline, his total mechanical energy is U + W1 + W2 = T = 84634 J. Since this is all kinetic energy , (1/2)*m*v^2 = T
v = sqrt(2T/m)
= sqrt(2* 84634/80) = 46.0 m/s.

When Sam reaches the bottom of the slope, he will be travelling at 46.0 m/s.

Answer 8

Took a long time to put this together. Didn't help that my answer got stuck for a while after clicking the Post button.