Newton's Laws
11. A 95.0 kg (209 lb) boxer has his first match in the Canal Zone (g = 9.782 m/s^2) and his second match at the North Pole (g = 9.832 m/s^2).
a. What is his mass in the Canal Zone?
ans. 95kg
b. What is his weight in the Canal Zone?
ans.
c. What is the mass at the North Pole
ans. 95kg
d. What is his weight at the North Pole?
ans.
e. Does he "weigh in" or does he really "mass-in"?
ans.
Mass-in. Sometime the word mass is used to replace the word weight.
12. Your new motorcycle weighs 2450 N. What is its mass in kilograms?
ans.
13. You place a 7.50kg television set on a spring scale. If the scale reads 78.4N, what is the acceleration of gravity at that location?
ans.
14. In Chapter 4 you calculated the braking acceleration for a car based on data in a drivers' handbook. The acceleration was -12.2 m/s^2.
If the car has a mass of 925kg, find the frictional force and state the direction.
ans.
Since the only acceleration is in the negative direction, the net force must be in that direction (Newton's 2nd Law). The only force acting in that direction is friction, so:
15. If you use a horizontal force of 30.0N to slide a 12.0kg wooden crate across a floor at a constant velocity, what is the coefficient of sliding friction between crate and floor?
ans.
We have not talked about frictional force yet, but you have enough knowledge to figure out how to calculate it. There is a property that materials have called coefficient of friction. The is represented by the Greek letter mu "m". The force caused by friction is:
To solve this problem, all we need to do is solve for m.
Notice that the coefficient of friction has no units.
16. You are driving a 2500kg car at a constant speed of 14.0 m/s along an icy, but straight and level, road. While approaching a traffic light, it turns red. You slam on the brakes. Your wheels lock, the tires begin skidding and the car slides t a halt in a distance of 25.0m. What is the coefficient of sliding friction (m) between your tires and the icy roadbed?
now we have mg = a. Now, since we do not know "a", we must use our kinematic equations to find "a".
Solve equation 1 for "a" and PLUG and CHUG!
Now plug this value back into the relationship that you found previously. (mg = a)
17. A person fishing hooks a 2.0 kg fish on a line that can only sustain a maximum of 38N of force before breaking. At one point while reeling in the fish, it fights back with a net force of 40N. What is the minimum acceleration with which he must play out line during this time in order to keep the line from breaking?
ans.
18. A 4500kg helicopter accelerates upward at 2 m/s^2. What lift force is exerted by the air on the propellers?
ans.
19. The maximum force a grocery sack can withstand and not rip is 250N. If 20 kg of groceries are lifted from the floor to a table with and acceleration of 5 m/s^2, will the sack hold?
ans.
OPPS! the sack rips, because the force is greater than 250N.
20. A student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads 836N.
a. As the elevator move up, the scale reading increases to 935N, then decreases back to 836N. Find the acceleration of the elevator.
ans.
b. As the elevator approaches the 74th floor, the scale reading drops as low as 782N. What is the acceleration of the elevator?
ans.
c. Using your results from parts a and b, explain which change in velocity, starting or stopping, would take the longer time.
ans.
Stopping, because the acceleration is less
d. Explain the changes in the scale you would expect on the ride back down.
ans.
As the elevator begins it decent "a" is negative and the scale reads less than 836N. When constant downward velocity is reached, the scale reads 836N since the acceleration in now zero. When the elevator is slowing at the bottom, the acceleration is positive and the scale reads more than 836N.
