Please show illustration and explain fully the problem below. 2) (4 pts each) A
ID: 1780065 • Letter: P
Question
Please show illustration and explain fully the problem below.
2) (4 pts each) A skier o? mass 50.0 kg skis dow:. to the bottom of an irregularly shaped hill. You don't know the exact shape of the hill, but you do know that tce hill is 35.0 m high vertically. She starts out moving with a speed of 10.0 m/s at the top of the hill Assuming there is no friction between the snow and skis and also no air resistance, predict the skier's speed at the bottom of the hill. Show your work and/or explain! a) In an actual experiment, the skier's speed at the bottom of the hill is measured to be 20.0 m/s. How much of the skier's initial mechanical energy (remember: mechanical energy potential energy) is lost in going from the top of the hill to the bottom of the hill? b) kinetic energy c) Where does this energy go? Show your work and/or explain! Wa(Fcos9)d Wnet.ext =AKE KE-2mw" PE,-mgy Closed SystemExplanation / Answer
a)
at the top of hill
velocity vi = 10 m/s
height h1 = 35 m
kinetic energy at the top kE1 = (1/2)*m*v1^2
potential energy PE1 = m*g*h1
total mechanical energy at the top E1 = kE1 + PE1
E1 = (1/2)*m*v1^2 + m*g*h2
at the bottom of hill
velocity vf = ?
height h2 = 0
kinetic energy at the bottom kE2 = (1/2)*m*v2^2
potential energy PE2 = m*g*h2
total mechanical energy at the bottom E2 = kE2 + PE2
E2 = (1/2)*m*v2^2 + m*g*h2
work done by internal dUint = 0
dKE + dPE + dUint = 0
(K2-K1) + (PE2-PE1) = 0
(1/2)*m*(v2^2 - v1^2) + m*g*(0-h1) = 0
v2 = sqrt(v1^2 + 2*g*h1)
v2 = sqrt(10^2 + (2*9.8*35))
v2 = 28 m/s <<<--------ANSWER
(b)
dKE + dPE + dUint = 0
(K2-K1) + (PE2-PE1) + dUint = 0
(1/2)*m*(v2^2 - v1^2) + m*g*(0-h1) + dUint = 0
(1/2)*50*(20^2-10^2) + 50*9.8*(0-35) + dUint = 0
dUint = 9650 J <<_---ANSWER
===============
(c)
this energy goes as waste in the form of heat
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.