Please I want someone to redo it clearly and step by step A 65-kg skier starts f
ID: 1596575 • Letter: P
Question
Please I want someone to redo it clearly and step by step
A 65-kg skier starts from rest and slides 120 m down a hill inclined at 30 degree to the horizontal. How much work does gravity do as the skier goes down the hill? How much work does the normal force do as the skier goes down the hill? For the previous problem, if the skier reaches the bottom of the hills travelling 30 m/s, what are The magnitude of the friction force acting on the skier? Use the work-KE theorem to solve The coefficient of sliding (kinetic) friction?Explanation / Answer
1)
a) Workdone by gravity, W_gravity = Fgx.d (here Fgx is the component of gravity along the inclined plane)
= Fgx*d*cos(0)
= m*g*sin(theta)*d
= 65*9.8*sin(30)*120
= 38220 J
b) Workdone by Nrmal force, W_norma = N*d*cos(90)
= 0
2)
a) Workdone by friction = change in mechanical energy
fk*d*cos(180) = (1/2)*m*v^2 - m*g*d*sin(30)
-fk = (1/2)*m*v^2/d - m*g*sin(30)
fk = m*g*sin(30) - (1/2)*m*v^2/d
= 65*9.8*sin(30) - (1/2)*65*30^2/120
= 74.7 N
b) fk = N*mue_k
mue_k = fk/N
= fk/(m*g*cos(30)
= 74.7/(65*9.8*cos(30))
= 0.135