1. Rubber tires and a wet blacktop have a coefficient of kinetic friction of = 0
ID: 1613261 • Letter: 1
Question
1. Rubber tires and a wet blacktop have a coefficient of kinetic friction of = 0.500. A pickup truck with mass 762 kg traveling 32.0 m/s skids to a stop. What is the horizontal acceleration of the truck?
2. An automobile with four -- wheel drive and a powerful engine has its mass of 1000 kg uniformly distributed on its four wheels. If the coefficient of static friction between the wheels and dry road is s = 0.80, starting from rest, what is the greatest forward horizontal acceleration the car can attain without spinning its wheels?
3. A 110 gram hockey puck sent sliding over ice is stopped in 15 m by the frictional force on it. If the puck had an initial speed of 6.0 m/s, what was the magnitude of the frictional force? {The kinetic frictional force IS Fnet,x }
4. In the above hockey puck problem, what was the coefficient of kinetic friction between the puck and the ice? {Ff = *FN = Friction Coefficient* Normal Force...}
5. A box weighing 143 N has a coefficient of static friction of 0.420 between the box and the floor.
6. What horizontal push must be applied to the box so that if any more force is applied it will start moving across the floor?
7. In the above problem with the 143-N crate, if that same push to just start the crate in motion is maintained after motion begins and the coefficient of kinetic friction between the bottom of the crate and the floor is 0.160, what will be the horizontal acceleration of the box?
8. In the above problem with the 143-N crate, what should the push be to move the box horizontally at constant velocity once motion begins?
9. A 500. kg sled is to be accelerated at a constant rate from rest by a horizontal constant force F to 525 km/h in 1.80 s. What is the magnitude of the net horizontal force needed to do this? (The net horizontal force is Fnet,x which includes F acting to the right and opposed by kinetic friction Ff acting to the left). {Since 1 N = 1 kg*m/s2, you MUST CONVERT km/h to m/s!}
10. Above you had a 500. kg sled being accelerated at a constant rate from rest by a horizontal constant force F to 525 km/h in 1.80 s. You found the net horizontal force needed to do this or Fnet,x which included F acting to the right and was opposed by kinetic friction Ff acting to the left. Now what is the magnitude of the kinetic frictional force Ff given that the pushing force F has a magnitude of 32200 N?
11. Above you had a 500. kg sled being accelerated at a constant rate from rest by a horizontal constant force F to 525 km/h in 1.80 s. You found the magnitude of the kinetic frictional force Ff giventhat the pushing force F has a magnitude of 32200. Now, what is the coefficient of kinetic friction k ? (Ff= k*FN = Friction Coefficient* Normal Force)
Explanation / Answer
1)
here by using the formula
a = u * g
a = 0.5 * 9.8
a = 4.9 m/s^2
3)
the magnitude of the friction force is
v^2 - u^2 = 2 * a * s
a = (v^2 - u^2) / 2*s
a = (0 - 6^2) / (2 * 15 )
a = -1.2 m/s^2
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.