Best of luck this semester!

]]>Part A = 3062 J

Solution Below:

In a certain library the first shelf is 15.0 cm off the ground, and the remaining four shelves are each spaced 31.0 cm above the previous one.

Part A

If the average book has a mass of 1.4 kg with a height of 22 cm, and an average shelf holds 29 books, how much work is required to fill all the shelves, assuming the books are all laying flat on the floor to start?

]]>Start by calculating the total mass for the books on a shelf. Since each shelf holds 29 books and each book is 1.4 kg, that gives a mass of 40.6 kg/shelf.

Now we can calculate the work required to fill each shelf. The work in this case just equals the energy done to move the books.

U = mgh

W = mgh

Now since the average book has a height of 22 cm, its center of mass is 11 cm above the base of the shelf. So 11 cm must be added to the height of each shelf. This gives the starting shelf a height of (0.11 + 0.15 = 0.26) 0.26 m. This is because the books were lying flat and have to be raised upright. So continue:

U = mgh

U = 71 * 9.8 * (0.26 + 0.57 + 0.88 + 1.19 + 1.5)

U = 71 * 9.8 * (4.4)

U = 3061.52 J

3062 J

Part A = 8.075 m/s

Part B = -0.255 m

Solution Below:

A 71 kg trampoline artist jumps vertically upward from the top of a platform with a speed of 4.7 m/s.

Part A

What is his speed as he lands on the trampoline, 2.2 m below his jump off point?

Start by figuring out how high the person will jump. Remember the kinematic equation v_{f}^2 = v_{i}^2 + 2ad. We know that the ending velocity (when he lands on the trampoline) will be zero, so:

v_{f}^2 = v_{i}^2 + 2ad

0^2 = 4.7^2 + 2 * (-9.8) * d

0 = 22.09 – 19.6d

22.09 = 19.6d

d = 1.127 m

So the person will reach a height of 1.127 m above the platform. Since the trampoline is 2.2 m below, they will fall a total of 3.327 m. We can calculate the potential energy and use this to figure out the person’s final speed:

U = mgh

U = 71 * 9.8 * 3.327

U = 2315

Convert to kinetic energy:

KE = 1/2mv^2

2315 = 1/2 * 71 * v^2

2315 = 35.5 * v^2

v^2 = 65.21

v = 8.075 ./s

8.075 m/s

Part B

If the trampoline behaves like a spring with spring stiffness constant 7.1 * 10^4 N/m , how far does he depress it? Any depression of the trampoline from equilibrium is to be taken as a negative distance.

]]>We found in Part A that the energy when the person hits the trampoline is 2315 J. The potential energy of a spring is 1/2kx^2, so we can solve:

U = 1/2kx^2

2315 = 1/2 * 71000 * x^2

0.065211 = x^2

x = 0.255 m

Remember that mastering physics says this is a negative distance – so the answer is -0.255 m

-0.255 m

Part A = 37 km

Solution Below:

You are flying to New York. You’ve been reading the in-flight magazine, which has an article about the physics of flying. You learned that the airflow over the wings creates a lift force that is always perpendicular to the wings. In level flight, the upward lift force exactly balances the downward weight force. The pilot comes on to say that, because of heavy traffic, the plane is going to circle the airport for a while. She says that you’ll maintain a speed of 300mph at an altitude of 20,000 ft. You start to wonder what the diameter of the plane’s circle around the airport is. You notice that the pilot has banked the plane so that the wings are 10° from horizontal. The safety card in the seatback pocket informs you that the plane’s wing span is 250 ft.

Part A

What is the diameter of the airplane’s path around the airport?

Express your answer with the appropriate units.

]]>Express your answer with the appropriate units.

This seems like a hard problem, but it’s actually fairly simple to solve. First of all, you know that the left is keeping the plane at the same altitude – so the lift (L) in the vertical direction must balance out gravity:

L_{y} = m * g

And you know that the vertical lift must also be:

L_{y} = L * cos(θ)

Since the vertical lift keeps the plane level, it must be equal to gravity. This means that:

L_{y} = m * g = L * cos(θ)

Therefore:

L = m * g / cos(θ)

Remember- we have to use cos here for the vertical lift, not sin. If this seems strange, think about what happens when the plane is turned 90° – the wings are pointing straight up and straight down, and there’s no way for the lift to keep the plane flying. It will crash. Anyway, since we know what the vertical lift is, we can also say that the horizontal lift is:

L_{x} = L * sin(θ)

This means:

L_{x} = L * sin(10)

L_{x} = L * 0.1736

This horizontal lift is what moves the plane in a circle. Since the plane is moving at 400 mph, we can plug this into the equation for centripetal force. We know the horizontal lift must equal the centripetal force, because the plane stays moving in a circle, without speeding up or slowing down:

L_{x} = m * a_{c}

L * 0.1736 = m * v^2 / r

Now, the problem gave us v, but we have to convert it from mph to m/s. One mph = 0.44704 m/s, so 400mph = 178.816 m/s

L * 0.1736 = m * 178.816^2 / r

Now substitute in for L:

(m * g / cos(θ)) * 0.1736 = m * 178.816^2 / r

The m cancels:

g / cos(θ) * 0.1736 = 178.816^2 / r

9.8 / cos(10) * 0.1736 = 178.816^2 / r

9.8 / 0.9848 * 0.1736 = 178.816^2 / r

9.95 * 0.1736 = 178.816^2 / r

1.7275 = 178.816^2 / r

1.7275 * r = 31975

r = 18509 m

Since the problem wants the diameter, just double the radius. This gives:

d = 37018 m

The problem seems to want the answer in kilometers, which is:

d = 37 km

37 km

Part A = 0

Part B = equal to

Part C = 500 kg·m/s

Solution Below:

A giant “egg” explodes as part of a fireworks display. The egg is at rest before the explosion, and after the explosion, it breaks into two pieces, with the masses indicated in the diagram, traveling in opposite directions.

Part A

What is the momentum PA,i of piece A before the explosion?

The problem tells us the egg is at rest before the explosion, so the momentum must be zero.

0

Part B

During the explosion, is the force of piece A on piece B greater than, less than, or equal to the force of piece B on piece A?

Remember that every action produces an equal and opposite reaction. So the two forces must be equal to each other.

equal to

Part C

The momentum of piece B is measured to be 500 kg·m/s after the explosion. Find the momentum PA,f of piece A after the explosion.

]]>Every action produces an equal and opposite reaction. Therefore, the momentums are equal. Piece A will be moving faster, since it has less mass, and piece B will be moving slower, since it has greater mass, but the momentums will be the same.

500 kg·m/s

Part A = F = -(μ * w) / ((μ * sin(θ)) – cos(θ))

Part B = 1/μ

**Solution Below:**

Consider a lawnmower of weight w which can slide across a horizontal surface with a coefficient of friction μ. In this problem the lawnmower is pushed using a massless handle, which makes an angle theta with the horizontal. Assume that F_{h}, the force exerted by the handle, is parallel to the handle.

Take the positive x direction to be to the right and the postive y direction to be upward.

Part A

Find the magnitude, F_{h}, of the force required to slide the lawnmower over the ground at constant speed by pushing the handle.

Express the required force in terms of given quantities.

Express the required force in terms of given quantities.

Since the lawnmower will be moving at constant speed, it won’t be accelerating. That means the frictional force must cancel out the force applied to move the lawnmower, so that the net force in the x-direction is zero. We can set up an expression to solve for this:

F * cos(θ) – F_{fr} = 0

Where F is the applied force, and F_{fr} is the frictional force.

Since the lawnmower isn’t moving in the y direction (it’s only moving horizontally across the ground), this means:

F * sin(θ) + w – F_{n} = 0

Where F_{n} is the normal force. Equivalently:

F_{n} = F * sin(θ) + w

We can simplify things by calculating frictional force in terms of the above:

F_{fr} = μ * F_{n}

F_{fr} = μ (* F * sin(θ) + w)

This may not seem obvious at first, but re-read what we did here: we equated the weight plus applied force in the y-direction to the normal force. By doing so, we were able to solve for the amount of frictional force. But we also know that the frictional force cancels out the applied force in the x-direction – so we can substitute in again:

F * cos(θ) = μ (* F * sin(θ) + w)

Having done this, we can rearrange the above to give:

F * cos(θ) = (μ * F * sin(θ)) + (μ * w)

(F * cos(θ)) – (μ * F * sin(θ)) = (μ * w)

F * (cos(θ) – (μ * sin(θ))) = (μ * w)

F = (μ * w) / (cos(θ) – (μ * sin(θ)))

Giving the answer:

F = (μ * w) / (cos(θ) – (μ * sin(θ)))

Note that Mastering Physics may want the following answer (it should take either, but just in case the above doesn’t work):

F = -(μ * w) / ((μ * sin(θ)) – cos(θ))

This is the same thing – it just swaps the order of the values in the denominator and adds a negative sign.

F = -(μ * w) / ((μ * sin(θ)) – cos(θ))

Part B

The solution for F_{h} has a singularity (that is, becomes infinitely large) at a certain angle θ_{critical}. For any angle θ > θ_{critical}, the expression for F_{h} will be negative. However, a negative applied force F_{h} would reverse the direction of friction acting on the lawnmower, and thus this is not a physically acceptable solution. In fact, the increased normal force at these large angles makes the force of friction too large to move the lawn mower at all.

]]>Find an expression for tan(θ_{critical})

The answer may not be obvious, but all the question is saying is that the force from friction will equal the force applied in the x direction. And since friction comes from the coefficient of friction (μ) multiplied by weight, we know that the force in the y-direction, multiplied by μ, must be the frictional force (which cancels out the force in the x-direction):

F * cos(θ_{critical}) = F * μ * sin(θ_{critical})

Cancel out the F from both sides:

cos(θ_{critical}) = μ * sin(θ_{critical})

Divide by cos:

1 = μ * sin(θ_{critical})/cos(θ_{critical})

Since tan = sin/cos, we can simplify:

1 = μ * tan(θ_{critical})

Now we can get the answer:

tan(θ_{critical}) = 1/μ

1/μ

**Solution Below:**

Part A

What must the charge (sign and magnitude) of a particle of mass 1.43 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 660 N/C?

Use 9.81 m/s^{2} for the magnitude of the acceleration due to gravity.

Use 9.81 m/s

For this problem, you just need to balance the force of gravity with the force from the magnetic field. The gravitational force is equal to (make sure to convert the mass of the particle to kilograms):

Fg = ma

Fg = 0.00143 kg * 9.81 m/s^{2}

Fg = 0.01403 N

The force from the magnetic field needs to be equal to the above, but in the opposite direction (so we need a force of -0.0140 N). The formula for the magnetic force is:

Fg = qE

We know the required force, and the magnitude of the field was already given:

-0.01403 N = q * 660

q = -2.125 * 10^-5 C

q = -2.13 * 10^-5 C

-2.13 * 10^-5 C

Part B

What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight?

Use 1.67 * 10^−27 kg for the mass of a proton, 1.60 * 10^−19 C for the magnitude of the charge on an electron, and 9.81 m/s^2 for the magnitude of the acceleration due to gravity.

]]>Use 1.67 * 10^−27 kg for the mass of a proton, 1.60 * 10^−19 C for the magnitude of the charge on an electron, and 9.81 m/s^2 for the magnitude of the acceleration due to gravity.

To solve this part, just use the same concept as in Part A:

Fg = qE

9.81 * m = q * E

9.81 * (1.67 * 10^−27) = (1.60 * 10^−19) * E

E = 1.0239 * 10^-7

E = 1.02 * 10^-7

E = 1.02 * 10^-7

Part A = 0.466 m

**Solution Below:**

Suppose a diode consists of a cylindrical cathode with a radius of 6.200*10^{-2} cm, mounted coaxially within a cylindrical anode with a radius of 0.5580 cm. The potential difference between the anode and cathode is 260 V. An electron leaves the surface of the cathode with zero initial speed (v_{initial} = 0).

Part A

Find its speed v_{final} when it strikes the anode.

Express your answer numerically in meters per second.

]]>Express your answer numerically in meters per second.

The size of the cathode and anode can be ignored – the electric potential is what matters here. Start by figuring out the energy difference due to the electric potential. This is equivalent to kinetic energy:

KE = qV

KE = 1.602 * 10^{-19} * 260

KE = 4.165 * 10^{-17}

Since KE = 1/2mv^2, set this formula equal to the above (note that the charge on an electron and the mass of an electron are just constants that you should memorize):

mv^2 = 8.330 * 10^{-17}

9.11 * 10^{-31} * v^2 = 8.330 * 10^{-17}

v^2 = 9.14 * 10^{13}

v = 9.16 * 10^{6} m/s

9.16 * 10^6 m/s

Part A = 0.466 m

**Solution Below:**

Two point charges, Q_{1} = −31μC and Q_{2} = 49μC, are separated by a distance of 12 cm. The electric field at the point P is zero.

Part A

How far from Q_{1} is P?

]]>The net electric field is given by the formula:

E(net) = (k * Q1 / r^{2}) + (k * Q2 / (r + x)^{2})

r is the distance from point P. The problem tells us that E(net) is 0, and x, the distance between the two charges, is 12cm (convert this to 0.12m):

0 = (k * Q1 / r^{2}) + (k * Q2 / (r + 0.12)^{2})

Subtract one of the two charges’ expressions (we used Q1):

-(k * Q1 / r^{2}) = (k * Q2 / (r + 0.12)^{2})

k cancels from both sides:

-Q1 / r^{2} = Q2 / (r + 0.12)^{2}

To simplify, multiply each side by the other’s denominator:

-Q1 * (r + 0.12)^{2} = Q2 * r^{2}

Divide by Q2:

(-Q1 / Q2) * (r + 0.12)^{2} = r^{2}

This gives us:

-((-31μC) / (49μC)) * (r + 0.12)^{2} = r^{2}

0.63265 * (r + 0.12)^{2} = r^{2}

Distribute out the (r + 0.12)^{2}:

0.63265 * (r^{2} + 0.24r + 0.0144) = r^{2}

This gives:

0.63265r^{2} + 0.152r + 0.0091 = r^{2}

Get everything onto one side:

0 = 0.36765r^{2} – 0.152r – 0.0091

Solve the quadratic equation:

r = 0.466 m

Part A = (a + gsin(θ) + μgcos(θ)) / (g – a)

**Solution Below:**

Block 1, of mass m_{1} = 0.650 kg , is connected over an ideal (massless and frictionless) pulley to block 2, of mass m_{2}, as shown. For an angle of θ = 30.0° and a coefficient of kinetic friction between block 2 and the plane of μ = 0.200, an acceleration of magnitude a = 0.450m/s^{2} is observed for block 2.

Part A

Find the mass of block 2, m_{2}.

]]>There are a couple ways to solve this problem. One method is to have a look at the problem Two Masses a Pulley and an Inclined Plane. The problem linked gives you a formula to find the ratio of the masses m1 and m2. This formula is:

m1/m2 = (a + gsin(θ) + μgcos(θ)) / (g – a)

This gives us:

0.650/m2 = ((0.450 + 9.8 * sin(30)) + (0.200 * 9.8 * cos(30))) / (9.8 – 0.450)

0.650/m2 = (2.205 + 1.70) / (9.350)

0.650/m2 = 0.4176

0.650 = 0.4176m2

m2 = 1.56 kg

1.56 kg