1. A dog is pulling a sled mass 200 kg at a steady speed of 4 m/s over ice with a
coefficient of friction of 0.3. The traces are making an upward angle of 30 degrees to the horizontal. The traces had a tension of 500 N. They break and the sled comes to rest. How long did this take?
2. A dog is pulling a sled mass 200 kg at a steady speed of 4 m/s over ice with a coefficient of friction (not given). The traces are making an upward angle of 30 degrees to the horizontal. The traces had a tension of 500 N. They break and the sled comes to rest. How long did this take?
The answer to number 2 is 1.62 seconds, but I really want to understand where that number came from. Thank you!
coefficient of friction of 0.3. The traces are making an upward angle of 30 degrees to the horizontal. The traces had a tension of 500 N. They break and the sled comes to rest. How long did this take?
2. A dog is pulling a sled mass 200 kg at a steady speed of 4 m/s over ice with a coefficient of friction (not given). The traces are making an upward angle of 30 degrees to the horizontal. The traces had a tension of 500 N. They break and the sled comes to rest. How long did this take?
The answer to number 2 is 1.62 seconds, but I really want to understand where that number came from. Thank you!
-
1)
a = µ*g = .3*9.8 = 2.94 m/s²
t = v/a = 4/2.94 = 1.36 sec
2)
µ = Ff/Fn = TcosΘ/m*g = 500*cos30°/(200*9.8) = .221
a = µ*g = 2.166 m/s²
t = v/a = 4/2.165 = 1.85 sec
Your answer is wrong
a = µ*g = .3*9.8 = 2.94 m/s²
t = v/a = 4/2.94 = 1.36 sec
2)
µ = Ff/Fn = TcosΘ/m*g = 500*cos30°/(200*9.8) = .221
a = µ*g = 2.166 m/s²
t = v/a = 4/2.165 = 1.85 sec
Your answer is wrong