A 10kg sled, initially at rest, is sliding down a snowy hill which makes an angl
ID: 2024059 • Letter: A
Question
A 10kg sled, initially at rest, is sliding down a snowy hill which makes an angle of 56° with respect to the horizontal. It starts at a height of 30m. Midway down the hill, the sled hits a patch of ice 10m long. If the ice is frictionless, and the coefficient of friction between the sled and snow is 0.08, determine the speed at the bottom of the hill.
I know this problem is divided into three parts, but I am unsure of how to even start this problem. The final speed SHOULD be 23.77 m/s, but I have no idea where to even begin.
Explanation / Answer
It's giving me an error drawing the force diagram. Taking the path of the slope to be the x-axis and the direction of the normal to be the y-axis.
Y-forces
Normal (+), Wcos (-)
Total Fy = may (ay=0 since it's not moving off the surface)
N-Wcos=0
N=Wcos=mgcos
X-forces
Friction (-), Wsin (+)
Total Fx = max
Wsin-Ffriction=max
mgsin-kN=max
mgsin-kmgcos=max
ax=g(sin-cos)
on snow
ax=9.81(sin56-0.08cos56)=7.694 m/s2
on ice
ax=9.81*sin56=8.133 m/s2
Need to determine the heights of the ice
h0=30 m
d0=distance on slope=30/sin56=36.18 m
h1=where ice begins=15 m
d1=distance left at midpoint=36.18/2=18.09 m
d2=distance where ice ends=18.09-10=8.09 m
h2=where ice ends=8.09*sin56=6.71 m
E0=E1 (Solving the equation for v1 to make things easier for the rest)
.5mv02+mgh0=.5mv12+mgh1
.5mv02+mgh0-mgh1=.5mv12
m(.5v02+g(h0-h1))=.5mv12
v12=v02+2g(h0-h1)
v1=(v02+2g(h0-h1))
v1=(02+2*9.81(30-15))=17.16 m/s
Now just do the same for the rest until you get vf
v2=(17.162+2*9.81(15-6.71))=21.38 m/s
vf=(21.382+2*9.81(6.71-0))=24.26 m/s
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