Posted by Dan on .
A snowmobile, with a mass of 530 kg, applied a force of 410 N backwards on the snow.
a) What force is responsible for the snowmobile’s resulting forward motion? (Hint: Think action–reaction force pairs.)
b) If the force of friction on the snowmobile is 187 N backwards, what is the net force acting on the snowmobile?
c)What is the acceleration of the snowmobile?
d)If the snowmobile accelerates for 7.3 s, what is its final speed?
e)How far would the snowmobile travel in this time?
f)The image above shows a person being “thrown” backwards as the snowmobile accelerates underneath him. Why would this happen?
A) we pushed back on the snow
Therefore Prof. Newton says (Law #3) that the snow pushes forward on us
B) 410 forward - 187 backward = 223 N forward
C) F = m a
223 = 530 (the driver evidently has no mass:) * a
a = 223/530 = .421 m/s^2
D) v = Vi + a t
v = 0 + .421 (7.3)
v = 3.07 m/s
E) average speed during acceleration = 3.07/2 = 1.54 so d = 1.54*7.3 = 11.2 m
(You could use d = (1/2) a t^2 but that is boring)
F) OH !! He has mass after all :)
The snowmobile accelerates
If the person is to accelerate with the snowmobile, there must be a forward force on the person.
If the seat is slippery and he rides with no hands, not forward force.
He did not accelerate. He stayed at the start hile the snowmobile took off.
A body of a mass 10.0kg sits at a distance f 1.5 metre from the pivot of a sea-saw. If another body of mass 20.0kg sits at a distance 1.0m from the pivot will the sea-saw balance horizontally