A 120lb person skis down a smooth parabolic incline Sec the

A 120-lb person skis down a smooth parabolic incline. (Sec the diagram.) Find the force exerted on the bottom of the skis by the snow when the skier passes the lowest point (at the origin). Neglect friction.

Solution

solution:

1)here body perform parabolic motion on curve as

x^2=4*K*y

from initial position we get that

4*k=256

x^2=256y

where y=x^2/256

to convert to meter we have to convert from feet to meter

y=x^2/78.02

2)to get accelaration at bottom we have to differentiate it withrespect time so we get

Vy=x*Vx/39.01 m/s

ay=Vx/39.01+x*ax/39.01 m/s^2

as at bottom velocity is maximum,hence

here by energy conservation we get

Vy=(2*g*h)^.5

h=100 ft=30.48 m

Vy at bottom is

Vy=24.45 m/s

where at x=160 ft=48.768 m

Vx=19.557 m/s

for ay=9.81 m/s^2

hence ax=7.446 m/s2

3)hence forces are along x and y direction are

F=533.78 N

m=54.41 kg or 120 lb

Fx=m*ax=54.41*7.446=405.136 N

Fy=m*ay=54.41*9.81=533.762 N

resultant force is

R=(Fx^2+Fy^2)^.5=670.1023 N at angle of 52.8 degree to horizontal.