1. Salim has a paper route. He puts his papers on a toboggan to pull them through the snow. The combined mass of the papers and the toboggan is 24kg. The coefficient of static friction between the toboggagn and the snow is 0.18. The coeefficient of kinetic friction between the toboggan and the snow is 0.10
a)with how much force must Salim pull in order to get the toboggan ro start moving?
b) how much force is needed to keep the toboggan moving at a constant speed?
2. When a car is accelerating, it will move in the direction of the net force. However, objects do not always move in the direction of the net force. Explain how this would be the case for a car coming to a stop at a traffic light? How would the diagram look with forces acting on the vehicle?
Find Fn by doing: Fn = mg, then, to find the minimum force required to start moving it, use Fk = Fn*Us ... (Us being static Friction).
To find the amount needed to keep it at a constant velocity,
use: Fs = Fn*Uk.
In your case, it would be:
Fk = 235.2*0.18 = 42.3N, to get it moving
Fs = 235.2*0.10 = 23.52, to keep it at a constant velocity.
1) a) To determine the force required to get the toboggan to start moving, we need to calculate the maximum static friction force. The formula for static friction is:
F(static friction) = coefficient of static friction * normal force
The normal force is equal to the weight of the toboggan and the papers, which is given by:
weight = mass * acceleration due to gravity
Given that the combined mass of the papers and the toboggan is 24 kg and the acceleration due to gravity is 9.8 m/s², we can calculate the weight:
weight = 24 kg * 9.8 m/s² = 235.2 N
Now we can calculate the maximum static friction force:
F(static friction) = 0.18 * 235.2 N = 42.46 N
Therefore, Salim must pull with a force greater than 42.46 N to get the toboggan to start moving.
b) To keep the toboggan moving at a constant speed, Salim needs to overcome the force of kinetic friction. The formula for kinetic friction is:
F(kinetic friction) = coefficient of kinetic friction * normal force
Using the same weight as before:
F(kinetic friction) = 0.10 * 235.2 N = 23.52 N
Therefore, Salim needs to pull with a force greater than 23.52 N to keep the toboggan moving at a constant speed.
2) When a car is coming to a stop at a traffic light, it experiences a deceleration, which means a net force acts in the direction opposite to its motion. The car's brakes apply a force that opposes its previous forward motion, resulting in a deceleration force.
The diagram representing the forces acting on the car would include the following:
- The force of friction acting in the opposite direction to the car's motion, caused by the brakes
- The weight of the car acting downward
- The normal force exerted by the road surface, equal in magnitude and opposite in direction to the weight
- Any other external forces acting on the car, such as air resistance or a gradient in the road's surface curvature
The net force acting on the car is equal to the difference between the force of friction and the weight of the car. As the car comes to a stop, the net force decreases until it reaches zero, at which point the car stops accelerating and moves at a constant speed (zero in this case).
To find the answers to these questions, we need to understand the concept of friction and Newton's laws of motion.
1. a) To find the force required to start the toboggan moving (the force of static friction), we can use the formula:
Fs = μs * m * g
where Fs is the force of static friction, μs is the coefficient of static friction, m is the mass, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the given values, we have:
Fs = 0.18 * 24 kg * 9.8 m/s^2
Fs ≈ 42.05 N
Therefore, Salim must pull with approximately 42.05 Newtons of force to start the toboggan moving.
b) To find the force needed to keep the toboggan moving at a constant speed (the force of kinetic friction), we can use the formula:
Fk = μk * m * g
where Fk is the force of kinetic friction and μk is the coefficient of kinetic friction.
Substituting the given values, we have:
Fk = 0.10 * 24 kg * 9.8 m/s^2
Fk ≈ 23.52 N
Therefore, a force of approximately 23.52 Newtons is needed to keep the toboggan moving at a constant speed.
2. When a car is accelerating, it moves in the direction of the net force acting on it. However, when a car is coming to a stop at a traffic light, the net force is acting in the opposite direction to its motion. Let's look at the forces acting on the car in this situation.
To stop the car, the brakes create a force called friction, which opposes the car's forward motion. This force of friction acts in the opposite direction to the motion of the car, creating a net force directed towards the back of the car (opposite to its original motion).
The diagram representing the forces acting on the car when coming to a stop at a traffic light would include:
- The force of friction, acting in the opposite direction to the car's initial motion.
- The weight or gravitational force acting downwards, which depends on the mass of the car.
- Other negligible forces such as air resistance or rolling resistance.
These forces combined create a net force in the opposite direction to the car's initial motion, which causes it to slow down and eventually come to a stop at the traffic light.
I think you can do #1.
#2)
F = m a
If your force is positive, your acceleration is positive. (negative mass does not occur in this subject)
HOWEVER
That says NOTHING about the magnitude or direction of VELOCITY.
If the car is headed North and puts the brakes on, the force will be South.
However until the speed reaches zero, the velocity remains North.