Sam, whose mass is 75 kg, straps on his skis and starts down a 50-m-high, 20 frictionless slope. A strong headwind exerts a horizontal force of 200 N on him as he skies. Use work and energy to find Sam’s speed at the bottom.

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;

The horizontal distance traveled by Sam is calculated as follows;

The work done by friction is calculated as follows;

Apply Work-Energy theorem to determine the speed of Sam at the bottom is calculated as follows;

Thus, the speed of Sam at the bottom of the inclined plane is 15.73 m/s.

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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

answer: question1

1,000 tons

question2

3: 8

question 3

store d

step-by-step explanation: