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
so
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
a) According to Newton's third law of motion, for every action, there is an equal and opposite reaction. Therefore, the force responsible for the snowmobile's resulting forward motion is the reaction force to the force it applied backwards on the snow. So, the forward motion force is also 410 N in the opposite direction.
b) Net force is the vector sum of all forces acting on an object. In this case, the force of friction on the snowmobile is 187 N backwards, and the force responsible for the forward motion is 410 N forwards. Subtracting the force of friction from the forward motion force gives the net force acting on the snowmobile.
Net force = Forward motion force - Frictional force
Net force = 410 N - 187 N
Net force = 223 N forwards
c) The acceleration of an object can be calculated using Newton's second law of motion:
Net force = mass x acceleration
Rearranging the equation to solve for acceleration:
Acceleration = Net force / mass
Acceleration = 223 N / 530 kg
Acceleration ≈ 0.421 m/s² forwards
d) The final velocity of an object undergoing uniform acceleration can be calculated using the equation:
Final velocity = Initial velocity + (acceleration x time)
Based on the problem statement, the initial velocity is not given, so we assume it to be zero.
Final velocity = 0 + (0.421 m/s² x 7.3 s)
Final velocity ≈ 3.07 m/s forwards
e) The distance traveled by an object undergoing uniform acceleration can be calculated using the equation:
Distance = (Initial velocity x time) + (0.5 x acceleration x time²)
Given that the initial velocity is zero, the equation simplifies to:
Distance = (0.5 x acceleration x time²)
Distance = 0.5 x 0.421 m/s² x (7.3 s)²
Distance ≈ 13.125 m forwards
f) The image shows a person being "thrown" backwards as the snowmobile accelerates underneath him because of Newton's third law of motion. The person exerts a reaction force on the snowmobile, which in turn exerts an equal and opposite reaction force on the person. This reaction force pushes the person backward, causing them to be "thrown" in that direction.
a) According to Newton's third law of motion, for every action, there is an equal and opposite reaction. In this case, the snowmobile applies a force of 410 N backwards on the snow. As per the third law, the snow exerts an equal and opposite force of 410 N forward on the snowmobile. Therefore, the force responsible for the snowmobile's resulting forward motion is 410 N.
b) The net force acting on the snowmobile is the sum of all the forces acting on it. In this case, the force of friction on the snowmobile is 187 N backwards. Since the applied force is 410 N backwards, we can calculate the net force by subtracting the force of friction from the applied force: 410 N - 187 N = 223 N. Therefore, the net force acting on the snowmobile is 223 N.
c) The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Using Newton's second law of motion, which states that force equals mass times acceleration (F = ma), we can rearrange the equation to solve for acceleration (a = F/m). Plugging in the values: acceleration = 223 N / 530 kg = 0.42 m/s². Therefore, the acceleration of the snowmobile is 0.42 m/s².
d) To find the final speed of an object undergoing constant acceleration, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity (assumed to be 0 in this case), a is the acceleration, and t is the time taken. Plugging in the values: v = 0 + (0.42 m/s²) * (7.3 s) = 3.066 m/s. Therefore, the snowmobile's final speed is 3.066 m/s.
e) The distance traveled by an object undergoing constant acceleration can be calculated using the equation s = ut + 0.5at², where s is the distance, u is the initial velocity, a is the acceleration, and t is the time taken. Since the initial velocity is assumed to be 0, the equation simplifies to s = 0.5at². Plugging in the values: s = 0.5 * (0.42 m/s²) * (7.3 s)² = 9.714 m. Therefore, the snowmobile would travel 9.714 meters in this time.
f) The image of a person being "thrown" backwards as the snowmobile accelerates underneath him is a result of Newton's third law of motion. As the snowmobile applies a force to the snow, the snow exerts an equal and opposite force on the snowmobile. According to Newton's third law, the person standing on the snowmobile experiences this same force applied in the opposite direction. Since the person and the snowmobile are not rigidly attached, the person experiences a force pushing them backwards, resulting in the "thrown" motion.