Ex. 3: A tall building casts a shadow that is 67.2 m long. The angle between the sun’s rays and the ground is θ = 50.0°. Determine the height of the building.

Ex. 4: A lakefront drops off gradually at an angle θ. At 14.0 m offshore the depth is 2.25 m. (a) What is the value of θ ? (b) What would be the depth d of the lake at a distance of 22.0 m from the shore?

Ex. 7: A displacement vector r has a magnitude of r = 175 m and points at an angle of 50.0° relative to the x axis. Find the x and y components of this vector.

CH 2

Ex. 5 - A parachute and brakes begin slowing a drag racer at t 0 = 9.0 s, the velocity is v 0 = +28 m/s. When t = 12.0 s, the velocity has been reduced to v = +13 m/s. What is the average acceleration of the dragster?

Kinematic equations v = v 0 + at x = vt = 1/2 (v 0 + v)t x = v 0 t + 1/2 at 2 v 2 = v ax

Ex. 6 - A speedboat has a constant acceleration of +2.0 m/s 2. If the initial velocity of the boat is +6.0 m/s, find its displacement after 8.0 seconds.

Ex. 7 - A jet is taking off the deck of an aircraft carrier. Starting from rest, the jet is catapulted with a constant acceleration of +31 m/s 2 along a straight line and reaches a velocity of +62 m/s. Find the displacement of the jet.

Ex 13 - A stone is dropped from rest from the top of a tall building. After 3.00 s of freefall, what is the displacement y of the stone?

Ex 14 - After 3.00 s of freefall, what is the velocity v of the stone in the previous example?

The acceleration due to gravity also affects the motion of upward moving objects, decreasing their velocity.

Ex 15 - A referee tosses a coin with an initial speed of 8.00 m/s. Neglecting air resistance, how high does the coin go above its point of release?

Ex 16 - What is the total time the coin is in the air?

The coin in the previous example begins with an upward velocity which decreases to zero, then increases in a downward direction, but the downward acceleration due to gravity is constant.

Velocity and acceleration can be plotted on a graph as displacement vs. time and as velocity vs. time.

In each case, the slope of the line indicates the velocity (rate of change of displacement) and acceleration (rate of change of velocity).

If an object is accelerating, a position vs. time graph will be a curved line. The velocity at any instant in time is the slope of a line tangent to the curve at that point.

The slope of the line of a velocity vs. time graph is the average acceleration.

CH 3

In projectile motion, the v x is constant (v x = v 0x ), a x = 0. But v y changes because of the acceleration due to gravity; a y = 9.80 m/s 2 usually.

Ex. 2 - An airplane flies horizontally with a constant velocity of +115 m/s at an altitude of 1050 m. The plane releases a “care package” that falls to the ground along a curved trajectory. Ignoring air resistance, determine the time required for the package to hit the ground.

Ex. 3 - Find the velocity of the package just before it hits the ground.

Ex. 5 - A placekicker kicks a football at an angle of θ = 40.0° above the horizontal axis. The initial speed of the ball is v 0 = 22 m/s. Find the maximum height H that the ball attains.

Ex. 6 - Determine the time of flight of the football between kickoff and landing.

Ex. 7 - Calculate the range R of the kickoff.

CH 4

Newton’s First Law - objects resist acceleration. Law of inertia. The mass of an object is a quantitative measure of inertia.

Newton’s second law - F = ma. The unit of force is the Newton. 1 Newton = 1kgm/s 2.

Newton’s Third Law - for every action (force) there is an equal and opposite reaction. The third law describes two different forces being applied to two different objects.

Free-body diagram A diagram that represents the object and the forces that act on it.

Newton’s Law of Universal Gravitation - every particle in the universe exerts a force on every other particle by the following equation: F = G m 1 m 2 /r 2 G = x Nm 2 /kg 2

Ex. 5 - What is the magnitude of the gravitational force that acts on each particle, assuming m 1 = 12 kg, m 2 = 25 kg, and r = 1.2 m?

Relation Between Mass and Weight - W = G M E m 2 /r 2 W = m g g is determined by G, ME, and r.

The normal force F N is one component of the force that a surface exerts on an object with which it is in contact, namely, the component that is perpendicular to the surface.

The forces on the block are balanced. Adding another downward force causes F N to increase. Adding an upward force, subtracts from W and F N decreases; possibly to zero.

Ex. 8 - In a circus balancing act, a woman performs a headstand on top of a man’s head. The woman weighs 490 N, and the man’s head and neck weigh 50 N. It is primarily the seventh cervical vertebra in the spine that supports all the weight above the shoulders. What is the normal force that this vertebra exerts on the neck and head of the man (a) before the act and (b) during the act?

Friction before an object starts to move is static friction. The maximum static frictional force, f SMAX, is the maximum frictional force that can be applied before motion and acceleration occur.

The ratio of f SMAX to F N is called the coefficient of static friction. It is represented by µ S. f SMAX /F N = µ S

f SMAX /F N = µ S This equation relates only the magnitudes of the vectors, not the directions.

Ex. 8 - A sled is resting on a horizontal patch of snow, and the coefficient of static friction is µ S = The sled and its rider have a total mass of 38.0 kg. Determine the horizontal force needed to start the sled barely moving.

When two surfaces are moving across each other, the friction is called kinetic friction f K. It is usually less than static friction. µ K is the coefficient of kinetic friction. f K /F N = µ K

Ex A sled is traveling at 4.00 m/s along a horizontal stretch of snow. The coefficient of kinetic friction is m K = How far does the sled go before stopping?

Tension - the force on a rope that would tend to pull the rope apart. A force on one end of a rope will be felt by the object to which it is attached. We will assume for our discussions that all ropes are massless.

Equilibrium - An object is in equilibrium when it has zero acceleration. The x-component of the net force and the y- component of the net force must both be zero. ∑F x = 0∑F y = 0

If an object is accelerating, these new equations apply: ∑F x = ma x ∑F y = ma y

Ex An 8500-kg truck is hauling a kg trailer along a level road. The acceleration is 0.78 m/s 2. Ignoring the retarding forces of friction and air resistance, determine (A) the magnitude of the tension in the horizontal drawbar between the trailer and the truck and (B) the force D that propels the truck forward.

Ex A flatbed truck is carrying a crate up a 10.00° hill. The coefficient of static friction between the truck bed and the crate is µs = Find the maximum acceleration that the truck can attain before the crate begins to slip backward relative to the truck.

Ex Block 1 (m 1 = 8.00 kg) is moving on a frictionless 30.0° incline. This block is connected to block 2 (m 2 = 22.0 kg) by a cord that passes over a massless and frictionless pulley. Find the acceleration of each block and the tension in the cord.

CH 5

In uniform circular motion the magnitude of the velocity vector is constant. The direction; however, is constantly changing. This change is an acceleration which is called “centripetal acceleration”.

a c = v 2 /r

The direction of the vector quantity a c is toward the center of the circle.

Ex. 3 - The bobsled track at the Olympics contained turns with radii of 33 m and 24 m. Find the centripetal acceleration for each turn for a speed of 34 m/s. Express the answers as multiples of g = 9.8 m/s 2.

An object in uniform circular motion is constantly being accelerated. This is a non- inertial frame of reference; and an object in uniform circular motion can never be in equilibrium.

The force that pulls an object into a circular path is called a centripetal force. F = ma, so F c = ma c ; therefore, F c = mv 2 /r

Ex. 5 - A model airplane has a mass of 0.90 kg and moves at a constant speed on a circle that is parallel to the ground. Find the tension T in the guideline (length = 17 m) for speeds of 19 and 38 m/s.

In vertical circular motion the centripetal force is the vector sum of the normal force and the component of the weight that pushes directly toward the center of the circle. In the lower half of the circular motion the centripetal force will be less (normal force minus weight) than in the upper half of the circle (normal force plus weight).

A the top of the circle, the centripetal force is normal force plus weight: F C = mv 2 /r = F N + mg At the correct speed, normal force can become zero and: F C = mv 2 /r = mg Solving the last two terms for v gives:v= √rg

At this speed, the track does not exert a normal force; mg provides all the centripetal force. The rider at this point experiences apparent weightlessness.

Ex A roller coaster loop has a radius of 10 meters. What is the minimum velocity required to keep the cars in the loop during the ride?

Ex An evil father is pushing his daughter in a swing. If he gleefully pushes her as hard as he can, what is the minimum velocity at which she will make a complete vertical circle on the swingset if the swing’s chain is 6 meters long?

Ex After a lengthy trial, the court decides that the punishment should fit the crime. The father is sentenced to be pushed in a circle on the same swingset. If he weighs six times as much as his daughter, what is his minimum speed to complete the circle? What other differences are there in the execution of his punishment?