PHY 101 Quiz #16 Solutions

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Physics 101
Quiz #16 Solutions
Vector Components
Given a vector A with magnitude A and angle to x-axis of 
Divide vector into components

• Ax = Acos
• Ay = Asin
Combine components back to vector

• Phythagoras’ Theorem
• A2 = Ax2 + Ay2
• tan = Ay/Ax
Vector Addition by Components

• C=A+B
• Cx = Ax + Bx
• Cy = Ay + By
• C = sqrt(Cx2 + Cy2)
Newton’s First Law / Inertia / Force Definition

• An object continues in a state of rest or in a state of motion at a constant speed along a
straight line, unless compelled to change that state by a net force
• Inertia is the natural tendency of an object to remain at rest or in motion at a constant speed
along a straight line
– Greater inertia of an object greater the net force required to change its velocity
• Mass of an object is a quantitative measure of inertia
• First law is redundant with second law
• What is purpose of the first law?
 Define force
• Push or pull that could potentially cause motion
 Define when First Law works
Newton’s Second Law

• A measure of motion is momentum = p = mv
• Rate of change of momentum = F
• F = p/t = (mv)/t = [(mv)f – (mv)i]/t
• F means sum of external forces
– Mass does not change, then

– F = ma
Newton’s Third Law

• When one body exerts a force on a second body, the second body exerts an oppositely
directed force of equal magnitude on the first body
Newton’s Law of Universal Gravity

• Every particle in the universe exerts an attractive force on every other particle
• For two particles that have mass m1 and m2 and are separated by a distance r, the force that
each exerts on the other is directed along the line joining the particles and has a magnitude
given by
• G = 6.673 x 10-11 Nm2/kg2
• Newton proved that an object of finite size can be considered a particle for purposes of the
gravitation law, provided the mass of the object is distributed with spherical symmetry about
its center
• In this case, r is the distance between the centers of the spheres and not between the outer
surfaces
• The gravitational forces that the spheres exert on each other are the same as if the entire mass
of each were concentrated at its center
Weight of Object at Surface of Earth
• m is the mass of the object

• ME is the mass of the Earth
• r is the distance from the center of the earth to the object
• r must be equal to or greater than the radius of the Earth
Normal Force
• Normal force is one component of the force that a surface exerts on an object with
which it is in contact
• This is component that is perpendicular to the surface
• Depending on the physical situation, the Normal Force can be vertical, horizontal, or
some other direction
Friction Force
• When an object is in contact with a surface, there is a force acting on the object
• Component of this force that is perpendicular to the surface is called the normal force
• When the object moves or attempts to move along the surface, there is also a
component of force that is parallel to the surface
• This parallel component is called the friction
Static Friction Magnitude

• When two surfaces are not in relative motion, the force of friction is called static
friction
• The magnitude fs of the static frictional force can have any value from zero up to a
maximum value of fsMax , depending on the applied force
• In this equation, ms is the coefficient of static friction, and N is the magnitude of the
normal force
Kinetic Friction Magnitude

• When two surfaces are in relative motion, the force of friction is called kinetic
friction
• Magnitude of the kinetic frictional force is given by fk
• fk = kN
• In this equation, k is the coefficient of kinetic friction, and N is the magnitude of the
normal force
Friction Force Direction

• Static Friction
 The direction is opposite to the direction in which the object would move if
there was no friction
• Kinetic Friction
 The direction is opposite to the direction in which the object is actually
moving
Uniform Circular Motion

Centripetal acceleration = v2/r
Centripetal force = mv2/r
Relationship between velocity and period: v = 2r/T
Work
F is force, d is distance moved, and  is angle between F and d.
Kinetic Energy
Potential Energy
or
Mechanical Energy
Work – Energy Theorem
Work Done by Gravity
Conservation of Energy
Mechanical energy is conserved when there are no non-conservative forces (friction)
Impulse (vector)
Momentum (vector)
means summation of the momentum over each individual object that makes up the
system.
Impulse-Momentum Theorem (vector)
Conservation of Momentum
Momentum is conserved when there are no external forces.
Momentum and Energy Conserved

Elastic collision
Object 1 hits object 2. Object 2 is initially not moving.
1. A 10-kg mass is held 1.0 m above a table for 25 s. How much work is done
during that period?
W = F(cos)s.
In this problem, there is no motion. Therefore, s = 0.
W = F(cos)(0) = 0 joules of work.
2. A 0.3 kg object slides 0.8 m along a horizontal tabletop. How much work is
done in overcoming friction between the object and the table if the coefficient
of friction is 0.20?
W= Fdcos
W = (mg)dcos180 = 0.2(0.3)(9.8)(0.8)(-1) = -0.47 J
Work against friction is 0.47 J
3. What is the work done by gravity when a 2.0-kg ball falls to the floor from a
height of 1.50 m? Is it positive or negative? Explain.
The force of gravity is the weight of the object.

F = weight = mg = 2.0(9.8) = 19.6 newtons
W = F(cos)d
The force is down and the direction of motion is also down so  = 0.
W = 19.6(1)(1.5) = 29.4 joules
The work is positive because the force and motion are in the same direction.
4. A 50-kg keg of beer slides upright down a 3.0 m-long plank leading from the
back of a truck 1.5 m high to the ground. Determine the amount of work
done on the keg by gravity.
Gravity is a conservative force. The work done by the force of gravity does not depend
on the path (the plank). The work only depend on the starting and ending points (the
vertical height of 1.5 m)
W = F(cos)d = mg(1)(1.5) = 50(9.8)(1)(1.5) = 735 joules
5. A college student earning some summer money pushes a lawn mower on a

level lawn with a constant force of 250 N at an angle of 300 downward from
the horizontal. How far does the student push the mower in doing 1.44 x 103 J
of work?
W = F(cos)d
1.44 x 103 = 250(cos30)d
d = 1.44 x 103/250/cos30 = 6.65 m
6. The net work done on an object moving along a closed path in a force field is
zero when it returns to the origin. Is this a conservative or nonconservative
force?
This is a conservative force.
For a conservative force, work only depends on the starting and ending points of the
motion. Work does not depend on the path moved. If the starting and ending points are
the same and work is zero (0), then the force is conservative.
7. If this book is placed on an ordinary table (there is friction) and slid along a
path that brings it back to where it started, is work done?
Work is done.
Friction is a nonconservative force. For a nonconservative force, work depends on the

path traveled. Even if the starting and ending points are the same, work is done because
there is a path traveled.
8. Suppose that a 0.149-kg baseball is traveling at 40.0 m/s.

(a) How much work must be done on the ball to stop it?
(b) If it’s brought to rest in 0.02 m, what average force must act on the
ball?
(a)
Use the Work-Energy Theorem.
W = KEf – KEi
When the ball stops its KEf is 0.
W = 0 – KEi = -½ mv2
W = - (.5)(0.149)(40.0)2 = -119.2 joules
(b)
We could use an equation of motion from Chapter 2, but we will use Work.
W = F(cos)d
F = W/(cos)d = -119.2/(1)/(.02 m) = -5955 newtons.
9. A child in a wagon is traveling at 10 m/s just as she reaches the bottom of a

hill and begins to climb a second hill. How high up the hill will she get before
the wagon stops, assuming negligible friction losses?
Energy is conserved in this problem, so we have

Ef =Ei
mghf + ½ mvf2 = mghi + ½ mvi2
Mass, m, cancels out of this problem.
hi = 0.0 m (at bottom of hill)

hf = ?
vf = 10 m/s
vi = 0 m/s (at the top of the hill, the wagon stops moving)
ghf + ½ (0)2 = 0 + ½ (10)2

hf = ½ (10)2/g = (0.5)(100)/9.8 = 5.1 m
10. A particle, starting from point A in the drawing, is projected down the
curved runway. Upon leaving the runway at point B, the particle is traveling
straight upward and reaches a height of 4.00 m above the floor before falling
back down. Ignoring friction and air resistance, find the speed of the particle
at point A.
This is a conservation of energy problem.
At the top of the rise after point B, the velocity is zero.

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Name: _________________________

Divide vector into components

Combine components back to vector

Vector Addition by Components

Newton’s First Law / Inertia / Force Definition

Newton’s Second Law

• F = p/t = (mv)/t = [(mv)f – (mv)i]/t

• F means sum of external forces

– Mass does not change, then

Newton’s Third Law

Newton’s Law of Universal Gravity

• G = 6.673 x 10-11 Nm2/kg2

Weight of Object at Surface of Earth

• m is the mass of the object

Static Friction Magnitude

Kinetic Friction Magnitude

Friction Force Direction

Uniform Circular Motion

F is force, d is distance moved, and  is angle between F and d.

Work – Energy Theorem

Work Done by Gravity

Impulse-Momentum Theorem (vector)

Momentum and Energy Conserved

W = F(cos)(0) = 0 joules of work.

The force of gravity is the weight of the object.

W = 19.6(1)(1.5) = 29.4 joules

W = F(cos)d = mg(1)(1.5) = 50(9.8)(1)(1.5) = 735 joules

5. A college student earning some summer money pushes a lawn mower on a

This is a conservative force.

Friction is a nonconservative force. For a nonconservative force, work depends on the

8. Suppose that a 0.149-kg baseball is traveling at 40.0 m/s.

9. A child in a wagon is traveling at 10 m/s just as she reaches the bottom of a

Energy is conserved in this problem, so we have

hi = 0.0 m (at bottom of hill)

ghf + ½ (0)2 = 0 + ½ (10)2

This is a conservation of energy problem.

At the top of the rise after point B, the velocity is zero.

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