Download Newtons Laws Practice Probs. 1. Calculate the force that must be

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.
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
+
-4N
=
In this example, 3N is the net force.
3N
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 = m g
• FW = (1 kg) (-10 m/s/s) = -10 N
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
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
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 =
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 =
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.
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
2
What will be the acceleration of a
sky diver when air resistance is
half the weight of the sky diver?
2
What will be the acceleration of a
sky diver when air resistance is
half the weight of the sky diver?
0.5g
Find:
A) The weight of the 1200 kg elevator.
B) The upward force on the elevator when it is
pulled upward at 2 m/s/s.
Find:
A) The weight of the 1200 kg elevator.
F = W = (1200 kg)(-10 m/s/s) = -12,000 N
B) 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
A 2 kg hanging block accelerates the
8 kg block on the frictionless table.
What is the acceleration of the two block
system?
A 2 kg hanging block accelerates the
8 kg block on the frictionless table.
A. What is the acceleration of the two
block system? a = 20 N/10 kg = 2
m/s/s
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?
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
Find the rate of acceleration of the 8 kg and 2 kg
blocks.
What is the tension in 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
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
The acceleration of
mg is down
(negative), while the
acceleration of Mg is
up (negative) as both
are connected via the
string.
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(-6 + 10) = 8 N
Note: g is NOT -10 as the sign is part of the eqtn.