1. Force: is a push or a pull.
There are contact forces such as pushing, pulling, or friction.
There are field forces such as gravitation and electrostatics.
Force is measured in Newtons, N.

2. Force, Mass, and Acceleration
Isaac Newton’s 2nd law of motion states: the acceleration of an object is directly proportional to the net force on an object and indirectly proportional to the acceleration of an object. That is,
a = Fnet / m
The net force, Fnet, is defined as the sum of all forces acting along a given line of action.
Example:
7N N = N
In this example, 3N is the net force.

3. Weight: is the attractive force between any mass and the mass of the earth.
Weight = mass x gravitational acceleration
FW = m g
where g = -10 m/s/s
So, the weight of 1 kilogram is -10 Newtons (N).
FW = (1 kg) (-10 m/s/s) = -10 N

4. Find: A) the weight of the block
Find: A) the weight of the block. B) the force applied to the block, if the acceleration of the block is 0 m/s/s. C) the net force on the block. D) the force applied to the block, if the acceleration of the block is 1.2 m/s/s. to the right.
F = ?
Friction = 5N
M = 10 kg

5. Find: A) the weight of the block
Find: A) the weight of the block. (W = 98 N) B) the force applied to the block, if the acceleration of the block is 0 m/s/s. (F = 5N) C) the net force on the block. (Fnet = 0 N) D) the force applied to the block, if the acceleration of the block is 1.2 m/s/s. to the right. (F = 17 N)
F = ?
Friction = 5N
M = 10 kg

6. Find A) the weight of the block
Find A) the weight of the block. B) the tension in the rope tied to the block, if the acceleration of the block is 1.2 m/s/s upwards.
T =
M = 10 kg
Weight =

7. Find A) the weight of the block
Find A) the weight of the block. (W = 98 N) B) the tension in the rope tied to the block, if the acceleration of the block is 1.2 m/s/s upwards. ( T = 110 N)
T =
M = 10 kg
Weight =

8. Newtons Laws Practice Probs. (accelerating objects) 1
Newtons Laws Practice Probs. (accelerating objects) 1. Calculate the force that must be applied to produce an acceleration of 1.8 g’s (18 m/s/s) on a 1.2 kg puck sliding free of friction on an air table.

9. Newtons Laws Practice Probs. 1
Newtons Laws Practice Probs. 1. Calculate the force that must be applied to produce an acceleration of 1.8 g’s (18 m/s/s) on a 1.2 kg puck sliding free of friction on an air table F = 21.6 N

10. 2 What will be the acceleration of a sky diver when air resistance is half the weight of the sky diver?

11. 2 What will be the acceleration of a sky diver when air resistance is half the weight of the sky diver? 0.5g

12. The weight of the 1200 kg elevator.
Find:
The weight of the 1200 kg elevator.
The upward force on the elevator when it is pulled upward at 2 m/s/s.

13. The upward force on the elevator when it is pulled upward at 2 m/s/s.
Find:
The weight of the 1200 kg elevator.
F = W = (1200 kg)(-10 m/s/s) = -12,000 N
The upward force on the elevator when it is pulled upward at 2 m/s/s.
Fnet = m a
F - 12,000 N = (1200 kg)(2 m/s/s)
F = 2,400 N +12, 000 N
F = 14, 400 N

14. A 2 kg hanging block accelerates the
8 kg block on the frictionless table.
What is the acceleration of the two block system?

15. A 2 kg hanging block accelerates the
8 kg block on the frictionless table.
What is the acceleration of the two block system? a = 20 N/10 kg = 2 m/s/s

16. What is the tension in the string supporting the 2 kg mass?
A 2 kg hanging block accelerates the
8 kg block on the frictionless table.
What is the tension in the string supporting the 2 kg mass?

17. A 2 kg hanging block accelerates the
8 kg block on the frictionless table.
The tension, T, in the string on the 2kg mass
F net = ma
T - (2 kg)(10 m/s/s) = (2 kg)(-2 m/s/s)
T = 20 N - 4 N
T = 16 N

18. Find the rate of acceleration of the 8 kg and 2 kg blocks.
What is the tension in the string?

19. Mg – T = Ma T – mg = ma Adding these yields: Mg – mg = (M + m) a
The acceleration of mg is down (negative), while the acceleration of Mg is up (negative) as both are connected via the string. T T mg Mg
Mg – T = Ma
T – mg = ma
Adding these yields:
Mg – mg = (M + m) a
a = g(M –m) / (M +m) so
a =- 10 (6/10) = -6 m/s/s

20. Note: g is NOT -10 as the sign is part of the eqtn.
The acceleration of mg is down (negative), while the acceleration of Mg is up (negative) as both are connected via the string. T T mg Mg
To find the tension in the string ( which is the same throughout the string), we may use:
T – mg = ma
T = ma + mg
T = m(a + g) = 2( ) = 8 N
Note: g is NOT -10 as the sign is part of the eqtn.