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?
force=mu*mg in both cases on 1.
1. To solve these problems, we need to understand the concept of static friction and kinetic friction. Static friction is the force that opposes motion between two objects that are not moving relative to each other. Kinetic friction, on the other hand, is the force that acts on objects in motion, opposing their motion.
a) To determine the force required to get the toboggan to start moving, we need to calculate the maximum static friction.
The formula for static friction is given by:
F_static = coefficient of static friction * normal force
In this case, the normal force is equal to the weight of the toboggan and papers, which can be calculated as:
Normal force = mass * gravitational acceleration
Plugging in the values, we have:
Normal force = 24 kg * 9.8 m/s^2 = 235.2 N
Now, we can calculate the maximum static friction:
F_static = 0.18 * 235.2 N = 42.3 N
Therefore, Salim must pull with a force of 42.3 Newtons to get the toboggan to start moving.
b) To determine the force needed to keep the toboggan moving at a constant speed, we need to calculate the kinetic friction.
The formula for kinetic friction is given by:
F_kinetic = coefficient of kinetic friction * normal force
Using the same normal force value as before, we have:
F_kinetic = 0.10 * 235.2 N = 23.52 N
Therefore, Salim needs to apply a force of 23.52 Newtons 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, it experiences a decelerating force, or negative acceleration. The force that slows down the car and eventually brings it to a stop is the braking force, which acts in the opposite direction of its motion.
In this case, the diagram showing forces acting on the vehicle would include three forces: the force of friction acting in the opposite direction of the car's motion, the normal force acting perpendicular to the surface of the road, and the weight of the car acting vertically downwards.
The force of friction opposes the car's motion and gradually brings it to a stop. As the car slows down, the force of friction decreases until it becomes zero when the car comes to a complete stop.