In the winter, you are playing with some kids who have sleds. The snow is fresh
ID: 1900319 • Letter: I
Question
In the winter, you are playing with some kids who have sleds. The snow is fresh and slippery enough that friction between the sleds and the snow can be neglected. You, however, have boots with good deep treads that don't slip on the snow. One sled has two kids on it, with a total mass (including the sled) of 68 kg. Starting from rest on level ground, you push on it horizontally with a constant force and get it up to a speed of 6.0 m/s after pushing for 4 seconds. What was the force (in newtons) that you applied to the sled while pushing on it? A different sled has just one kid on it, with a total mass of 48 kg. Starting from rest on level ground, you push on it horizontally with the same amount of force that you used with the first sled. How long does it take to get this sled up to a speed of 6.0 m/s ? Later, you push the second sled (with the one kid) again with the same force, but this time it's on an area of rough, dirty snow that applies a constant friction force of 50 N opposite the direction the sled is moving. Now how long does it take you to get this sled up to a speed of 6.0 m/s, starting from rest? Once the sled from part (c) is up to 6.0 m/s, you suddenly stop pushing. How far does it travel from where you stop pushing to where it comes to rest? Assume that it stays on similar rough snow.Explanation / Answer
a)a=v/t=6/4=1.5m/s2
F=ma=68*1.5=102N
b)a=F/m=102/48=2.125m/s2
by first equation of motion,
v=0+at
t=v/a=2.82s
c)a=(F-friction)/48=1.083m/s2
t=6/1.083=5.54s
d)Acceleration after pushing has been stopped=-50/48=-1.0417m/s2
By third eqn. of motion
0-62=2*-1.0417*s
s=17.28s