Light Problem (Mirrors and Lenses)

A candle with a flame 1.5cm tall is placed 5.0cm from the front of a concave mirror.  A virtual image is produced that is 10cm from the vertex of the mirror. (a) Find the focal length and radius of curvature of the mirror (b) How tall is the image of the flame?

Energy Problem (The Hiker)

To calculate the climber's change of energy of gravity, the Ug=mgh equation can be used.  The mass is given as 82kg and gravity is 9.8m/s.  The height is changing by 540meters during the two hour period.  By multiplying them all together, the change in energy due to gravity is 433944Joules.

The second part of the question asks to find the power needed to increase the hikers Eg.  Here, the power equation can be used which is Power equals work divided by time.  The work is the change in energy just calculated, 433944Joules.  The time is 2hours which needs to be converted to seconds to be calculated properly.  One hour contains 3600seconds.  Therefore, 3600seconds multiplied by 2hours equals a time of 7200seconds.  Finally to find the power, take the work, 433944Joules, divided by the time in seconds, 7200seconds, to find the power generated which is 60.27W.  60.27W represents the power generated to increase the hiker's Eg.

Energy Problem (The Roller Coaster)

A roller coaster travels on a frictionless track as shown in Fig 5.31. (a) If the speed of the roller coaster at point A is 5.0m/s, what is its speed at point B? (b) Will it reach point C? (c) What minimum speed at point A is required for the roller coaster to reach point C?

To create the bar graphs, the knowledge of what K, Ug, and Us is in order.  While Ug represents the height of the object, K symbolizes the potential energy and Us represents a spring energy.  During point A, the initial energy graph needs an equal amount of K and Ug.  This is because the roller coaster is 5.0meters off the ground and therefore has the potential energy, Ug, to go down the hill.  There is no spring throughout this ride.  At point B, the intermediate energy the height, Ug, is gone and transferred to potential energy, K.  Since the roller coaster started out with 5.0m/s velocity and at the top of a hill, the coaster now needs all the energy it lost from its height, Ug, in potential energy to make it up the second hill.  By point C, the final energy graph represents the transfer of all the energy to Ug, height.  The coaster is now at 8.0 meters in height, which is taller than the first hill.  The potential energy it had at points A and B were used to propel the roller coaster to point C.

The speed at point B can be calculated with the equations K, potential energy, equals 1/2 mass multiplied by velocity squared and Ug, height, equals mass multiplied by gravity, 9.8m/s, and height, in meters.  Since the mass of the coaster isn't given, the combination of the equations need to be used to find the speed at point B.  If 1/2 mass times velocity squared plus mass times gravity times height is put equal to 1/2 mass times velocity at point B squared, the velocity can be determined.  Because the mass isn't given, one can begin by dividing the entire equation by mass.  Then, plug in the numbers-1/2 (5m/s) squared + (9.8m/s) (5m) equals 1/2 B's velocity squared.  B's velocity is then found to be 11.09m/s.

To find if the coaster reaches point C, the right side of the equation above is used equaled to the height equation, Ug equals gravity times height times mass.  Again, divide the entire equation by mass.  Plug in the numbers, 1/2 (5m/s) squared + (9.8m/s) (5m) equals (9.8m/s) times height.  By solving for height, the height comes out to be 6.28m.  The coaster doesn't reach point C at 8meters.

Finally to locate the minimum speed, velocity, required to reach point C from point A, use the equation from the last paragraph keeping the velocity as the variable, 1/2(v) squared + (9.8) (5m) ='s (9.8) (8).  The height of A is used in the first half of the equation because the coaster starts here and ends at 8meters.  That's why 8meters is used in the second half of the equation. The velocity needed at point A to reach 8meters at point C is 7.67m/s.

Circular Motion Problem

The centripedal force is defined as the mass times the velocity squared divided by the radius of the loop.  The mass of the person is 55kg and her velocity she is traveling by is 65m/s while the radius of the loop is 380m when plugged into the formula the force is 612N.  612N is the force that is pushing the airplane out towards the rim of the loop.

The weight or gravity force is always and can be calculated by multiplying the mass of the women by the gravity force.  The centripetal force calculated before is bigger at 612N, because the centripetal is larger antoher force is therefore normal force.  Since the weight and centripetal forces should equal each other in order to keep the rider from being thrown off the plane.  The Fn is calculated using math skills.  Fa plus Fn must equal 612N. Fo equals 539N.  Using math Fn is found to be 73N.  This 73N force is most likely provided by a protector or harness or any other mechanism keeping the person strapped in.  This is balancing the forces.  The Fn is downward along with the weight force in order to create the state of equilibrium.

The woman is pushed out with 612N when it's greater than the persons weight force pull (539N) because of its greater force exerted outwards.  The person feels lighter than usual.

The Fl represents the normal force on the human.  The Fn or W is the weight force of the human while in the air.  The Fl arrow is straight up to counteract the weight force or Fn arrow.  The two forces need to unbalance or in this case be different because the human is in motion at the time.

Momentum (Collision) Problem

First, grams needs to be converted to kg so the 15g is divided by 1000g which is one kilo.  The 15grams is then converted to 0.015kg.

To find the change in momentum, mass is multiplied by change in velocity.  The velocity is given as 600m/s and the mass was just calculated as 0.015kg.  The change in momentum is found to be 9N.m.  This is the change of the first vehicle.

Next, the other car is 100grams and should be divided by 1000grams which is one kilo to find out the car is 0.1kg in mass.

The change of velocity of both cars is next found with the equation change in momentum equals mass times change in velocity.  The change of momentum was found as 9N.m.  The two masses are added together to find the change of velocity form both cars.  Added together, the mass of 0.115kg is divided by the momentum of both cars.  Finally, the change of velocity in both cars is found as 78.26m/s.

Momentum (Impulse Problem)

The equation of change in momentum equals force multiplied by change in time is implemented.  The problem gives 1320N of net force is acting on the ball.  It also gives 9.0 times 10 to the negative third power seconds as the change in time the ball goes.  By multiplying the Force and change of time together, the momentum is calculated as 11.88N/s.

The to find the change in velocity, the equation change in momentum equals mass multiplied by change in velocity.  The momentum was found to be 11.88N/s, and earlier the mass is given as 140g.  Since the mass is needed in Kg, one will  need to divide 140g by 1000grams because that is 1 Kilo.  The mass will then equal 0.140 kg.  With the mass, one is finally able to divide the momentum of 11.88N/s by the mass, 0.140kg, to find the change in velocity which is 84.86m/s.

Contact Forces Problem

A crate resting on a floor has weight, normal, friction, and Force forces.  The normal and weight forces are balanced.  The friction and Force forces aren't balanced because the object is moving forward therefore there is a positive force to the right.  Friction is negative because it takes away force by slowing it down.

To find the weight force, mass multiplied by acceleration is enforced (ma=W).  The weight force is negative because acceleration is negative (-9.8m/s2).

Normal force and weight force are balanced, so the normal will equal the weight force except a positive number.  The normal and weight forces added together should equal zero.

Next, the frictional force should be calculated using the friction equals mu multiplied by Normal equation. N signifies normal force, while M represents the coefficient of friction.  The coefficient of friction is given (0.15) and is mulitplied by the Normal just calculated (490N) to find friction (73.5N).

Finally to calculate acceleration, the equation Net Force equals mass multiplied by acceleration should be used.  EF signifies net force which will be Force subtracted by friction.  The mass is given in the problem (50kg).  To solve for acceleration, subtract the friction force from the Force.  Then, divide that number by the mass to find acceleration.

Acceleration Problem

Since the displacement is less than 13 meters, the driver will not hit the child given a reaction time of 0.25 seconds.

Conversions:
Since the problem gives us the speed limit in 40km divided by one hour times 1 hour over 3600 seconds times 1000 meters over 1 kilometer gives us 11.1 m/s.

xt graph shape:
The problem states the drivers reaction time is 0.25 seconds.  Therefore, the driver is moving at constant speed for the first 0.25 seconds of the graph.  At 0.25 seconds, the driver reacts to the child in the street and begins to slow his car by breaking.  Since time is still moving forward and also the car is moving forward, the graph has a y squared equals mx plus b equation.

vt graph shape:
In the first 0.25 seconds of the graph, the car is moving at 11.1 meters per second forward.  Since the car is moving at a constant velocity forward, this correlates into a horizontal line on the vt graph.  At 0.25 seconds, the driver reacts and begins to break.  His acceleration is -8 meters per a second squared.  On the graph, there is a downward linear line to reciprocate the driver breaking until stopping.

at graph shape:
In the first 0.25 seconds, the car is moving forward at a constant velocity.  Since the car is not accelerating at this moment, the graph begins at t equals zero for 0.25 seconds.  After 0.25 seconds, the car begins to negatively accelerate.  This is represented on the graph by a horizontal line in the negative quadrant.  The line is at -8 meters per second squared, because that is the drivers acceleration.  The driver finally comes to a stop represented by the time returning to t equals zero.

A equals change in velocity over change in time explanation:
The problem wants to figure out if the automobile is able to stop before hitting the child.  The child is at 13 meters.  Since the acceleration of the vehicle and the initial velocity are given in the problem, I can use these to find the amount of time the vehicle takes to stop.  The vehicle stops after 1.387 seconds after applying the brake.

A equals lw+1/2bh:
After I find the change in time the driver took, I am able to find the displacement under the line of the vt graph to find the displacement of the automobile.  I can further divide the area up into a rectangle and a triangle.  Using the formulas length times width for a rectangle and 1/2 base times height for a triangle, I will be able to find the area which is the displacement by adding them together.

lw (the rectangle) explanation:
The length of the rectangle in the vt graph is the intial velocity the vehicle is going at.  The width is the amount of time (reaction time) 0.25 seconds the driver takes to realize to begin to brake.  The area of the rectangle is the displacement of the reaction time of the driver.

1/2bh explanation:
The driver begins to brake forming a triangle on the vt graph of the second half of the graph.  I found the change in time earlier with acceleration equals change in velocity over change in time of the vehicle.  I multiply all these together times 1/2 to find the displacement of the vehicle.

Solution and Conclution:
After adding the areas of the triangle and rectangle under the velocity time graph line, the distance the car went is produced.  The car went 10.5 meters away from the car.  The child was safe from the car by 2.5 meters.

Projectile Motion Problem

The y-graph represents the problem, because the car begins at above the origen then is thrown to the Earth after a certain amount of time goes by.  The slope is always negative, so the derivative and Vy graph is a negative linear line.  As time goes by, the position approaches 0.  As it approaches 0, it approaches 0 faster.  The object starts at rest and begins going faster and faster in a negative direction.

The x-graph represents the problem, because as time goes on the car is still horizontally moving.  The derivative of a positive linear line the poitive position go in a positive direction at a positive rate.

Equations:
First, I used the position equation.  The car travels from 0 meters to 125 meters, so the position is -125 meters.  The vertical motion doesn't have an initial velocity.  Gravity pulls on the car at -9.8 m/s2.  I then solve the equation to find the time the car took to reach the ground.  I find after 5.05 seconds, the tornado replaces the car on the ground.

Next, I use the change in x equation, change in x equals Vx times t, to find how far horizontally the car actually travels.  The tornado flings the car horizontally at an initial velocity of 90.0m/s.  Since I already found the time the car takes to reach the ground, I can plug 5.05 seconds in for time.  I come up with the horizontal displacement of the car by the tornado to be 454.5 meters.