Answer:
Explanation:
Sam mass=75kg
Height is 50m
20° frictionless slope
Horizontal force on Sam is 200N
According to the work energy theorem, the net work done on Sam will be equal to his change in kinetic energy.
Therefore
Wg - Ww =∆K.E
Note initial the body was at rest at top of the slope.
Then, ∆K.E is K.E(final) - K.E(initial)
K.E Is given as ½mv²
Since initial velocity is zero then, K.E(initial ) is zero
Therefore, ∆K.E=½mVf²
Wg is work done by gravity and it is given by using P.E formulas
Wg=mgh
Wg=75×9.8×50
Wg=36750J
Ww is work done by wind and it's is given by using formulae for work
Work=force × distance
Ww=horizontal force × horizontal distance
Using Trig.
TanX=opposite/adjacent
Tan20=h/x
x=h/tan20
x=50/tan20
x=137.37m
Then,
Ww=F×x
Ww=200×137.37
We=27474J
Now applying the formula
Wg - Ww =∆K.E
36750 - 27474 =½×75×Vf²
9276=37.5Vf²
Vf²=9275/37.5
Vf²= 247.36
Vf=√247.36
Vf=15.73m/s
The speed of Sam at the bottom of the inclined plane is 15.73 m/s.
The given parameters;
The gravitational potential of Sam is calculated as follows;
P.E = mgh
P.E = 75 x 9.8 x 50
P.E = 36,750 J
The work done by friction is calculated as follows;
[tex]W_x = F \times d[/tex]
The horizontal distance traveled by Sam is calculated as follows;
[tex]tan\ 20 = \frac{50}{d} \\\\d = \frac{50}{tan \ 20} \\\\d = 137.36 \ m[/tex]
The work done by friction is calculated as follows;
[tex]W_x = 200 \times 137.36\\\\W_x = 27,472 \ J[/tex]
Apply Work-Energy theorem to determine the speed of Sam at the bottom is calculated as follows;
[tex]\Delta W = \Delta K.E \\\\(36750 - 27,472) = \frac{1}{2} \times 75\times (v_f^2-0^2)\\\\9278 = 37.5v_f^2\\\\v_f^2 = \frac{9278}{37.5} \\\\v_f^2 =247.4\\\\v_f = \sqrt{247.4} \\\\v_f = 15.73\ m/s[/tex]
Thus, the speed of Sam at the bottom of the inclined plane is 15.73 m/s.
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