Mechanics Exam - AP Physics C: Electricity and Magnetism

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Question

A guillotine blade weighing is accelerated upward into position at a rate of $2.0\frac{m}{s^2}$ .

What is the the approximate mass of the guillotine blade?

Answer

The force of gravity on the blade is , which is the same as $400\frac{kg\cdot m}{s^2}$

This unit relationship comes from Newton's second law.

is the mathematical expression of Newton's second law. The units for force must be a product of the units for mass and the units for acceleration.

Solve the expression by plugging in known values.

$400 \frac{kg\cdot m}{s^2}=m \cdot 9.8\frac{m}{s^2}$

$\frac{400}{9.8}kg=mass\approx41kg$

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Question

A guillotine blade weighing is accelerated upward into position at a rate of $2.0\frac{m}{s^2}$ .

What is the tension on the rope pulling the blade, while it is accelerating into position?

Answer

The tension in the rope is the sum of the forces acting on it. If one considers that the net force on an object must equal the mass of the object times the acceleration of the object, the net force on the object must be the force due to tension from the rope minus the force due to gravity.

$F_{T}-F_{g}=ma$

Rearrange the equation.

$F_{T}=ma+F_{g}$

Plug in known values.

$F_T=41kg\cdot2\frac {m}{s^2} + 400N$

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Question

An object is moving in two dimensions. Its vertical motion relative to the horizontal motion is described by the equation $\small y=ax^2+bx +c$ . Its motion in the horizontal direction is described by the equation $\small x=v_{x}t$ . What is the object's velocity is the direction in terms of its horizontal position $\small x$ ?

Answer

The y velocity is thetime derivative of the $\small y$ position, and not the $\small x$ derivative. In order to find it, use the chain rule:

$\frac{dy}{dt}=\left(\frac{dy}{dx}\right) \frac{dx}{dt}$

$\frac{dy}{dx}= 2ax + b$

Of course, $\frac{dx}{dt} =v_{x}$

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Question

Two objects moving in one dimension created the following velocity vs. time graph:

From the graph above, what is true about the two objects at time $\small t=5s$ ?

Answer

Since this is a graph of velocity and not position, the curves intersect where the velocities match. Since we do not know the starting position, we do not know where the objects are relative to one another.

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Question

Two objects moving in one dimension created the following velocity vs. time graph:

From the graph above, which object has traveled a greater distance from its starting position when $\small t=5s$ ?

Answer

Since this is a graph of velocity vs. time, its integral is distance travelled. We can estimate the integral by looking at the area under the curves. Since Bbject 1 has a greater area under its velocity curve, it has covered a greater distance. Its velocity is greater that Object 2's for the entire time, so it makes sense that it will travel farther.

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Question

Atwood's machine consists of two blocks connected by a string connected over a
pulley as shown. What is the acceleration of the blocks if their masses are and .

Assume the pulley has negligible mass and friction.

Answer

From the force diagram above, we can see that tension is pulling up on both sides of the string and gravity is pulling down on both blocks. With this information we can write 2 force equations:

If we add the two equations together, we get:

where

Solving for , we get

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Question

A train car with a mass of 2400 kg starts from rest at the top of a 150 meter-high hill. What will its velocity be when it reaches the bottom of the hill, assuming that the bottom of the hill is the reference level.

Answer

The law of conservation of energy states:

If the car starts at rest, then the initial kinetic energy = 0 J.

If the car ends at the reference height, the final potential energy = 0 J.

Subsituting these values, the equation becomes:

The initial potential energy can be determined by:

$PE_i = mgh_i \ PE_i = 2400 kg \cdot 9.81 \frac{m}{s^2} \cdot 150 m\ PE_i = 3531600 \frac{kg \cdot m^2}{s^2}\ PE_i = 3531600 J$

The final kinetic energy equation is:

$KE_f = \frac{1}{2}mv_f^2\ KE_f = \frac{1}{2}(2400 kg)v_f^2$

Substituting the initial potential energy and final kinetic energy into our modified conservation of energy equation, we get:

$3531600 J = \frac{1}{2}(2400 kg)v_f^2\ 3531600 \frac{kg\cdot m^2}{s^2} = 1200 kg \cdot v_f^2\ \frac{3531600 \frac{kg\cdot m^2}{s^2}}{1200 kg}=v_f^2\ 2943 \frac{m^2}{s^2}=v_f^2\ \sqrt{2943 \frac{m^2}{s^2}}=\sqrt{v_f^2}\ 54.25 \frac{m}{s}=v_f$

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Question

A 120kg box has a kinetic energy of 2300J. What is its velocity?

Answer

The formula for kinetic energy of an object is:

$KE=\frac{1}{2}mv^2$

The problem gives us the mass and the kinetic energy, and asks for the velocity, so we can rearrange the equation:

$v=\sqrt{\frac{2KE}{m}}$

Use our given values for kinetic energy and mass to solve:

$v=\sqrt{\frac{2(2300J)}{120kg}}=6.2\frac{m}{s}$

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Question

An object has a mass of 5kg and has a position described by the given function:

What is the object's kinetic energy after two seconds?

Answer

Kinetic energy is defined by the equation:

$KE=\frac{1}{2}mv^2$

Taking the derivative of the position function allows us to obtain the velocity function:

We can now determine the velocity after two seconds:

Now that we know our velocity, we can solve for the kinetic energy.

$KE=\frac{1}{2}(5kg)(3\frac{m}{s})^2=22.5J$

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Question

An object starts from rest and accelerates at a rate of $2.5\frac{m}{s^2}$ . If the object has a mass of 10kg, what is its kinetic energy after three seconds?

Answer

Kinetic energy is given by the equation:

$KE=\frac{1}{2}mv^2$

We can find the velocity using the given acceleration and time:

$v=v_o+a\Delta t$

$v=0\frac{m}{s}+(2.5\frac{m}{s^2})(3s)=7.5\frac{m}{s}$

Use this velocity to find the kinetic energy after three seconds:

$KE=\frac{1}{2}(10kg)(7.5\frac{m}{s})^2=281.25J$

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Question

An object starts from rest and accelerates to an angular velocity of $12\frac{rad}{s}$ in three seconds under a constant torque of $50\ N\cdot m$ . How many revolutions has the object made in this time?

Answer

Since it is experiencing a constant torque and constant angular acceleration, the angular displacement can be calculated using:

$\Delta \theta =\omega_ot+\frac{1}{2}\alpha t^2$

The angular acceleration is easily calculated using the angular velocity and the time:

$\alpha=\frac{\Delta\omega}{\Delta t}=\frac{12\frac{rad}{s}-0\frac{rad}{s}}{3s}=4\frac{rad}{s^2}$

Using this value, we can find the angular displacement:

$\Delta \theta =(0\frac{rad}{s})(3s)+\frac{1}{2}(4\frac{rad}{s^2})(3s)^2$

$\Delta \theta=18rad$

Convert the angular displacement to revolutions by diving by $2\pi$ :

$\frac{18rad}{2\pi\frac{rad}{rev}}\approx2.9rev$

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Question

A circular disk of radius 0.5m and mass 3kg has a force of 25N exerted perpendicular to its edge, causing it to spin. What is the angular acceleration of the disk?

Answer

We can find the angular acceleration using the rotaional motion equivalent of Newton's second law. In rotational motion, torque is the product of moment of inertia and angular acceleration:

$\tau = I\alpha$

The moment of inertia for a circular disk is:

$I=\frac{1}{2}mr^2$

The tourque is the product of force and distance (in this case, the radius):

$\tau=F\cdot r$

We can plug these into our first equation:

$(F)(r)=(\frac{1}{2}mr^2)(\alpha)$

Simplify and rearrange to derive an equation for angular acceleration:

$\alpha=\frac{Fr}{\frac{1}{2}mr^2}=\frac{2F}{mr}$

Use our given values to solve:

$\alpha=\frac{2(25N)}{(3kg)(0.5m)}=33.3\frac{rad}{s^2}$

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Question

A 0.18 m long wrench is used to turn the nut on the end of a bolt. A force of 85 N is applied downward to the end of the wrench, as shown in the figure. The angle between the force and the handle of the wrench is 65 degrees.

What is the magnitude and direction of the torque (around the center of the bolt) due to this force?

Answer

To calculate the magnitude of the torque,

$\tau =rF\sin\theta$

where the radius is the distance between the center of rotation and the location of the force and the angle $\theta$ is between the radius and the force .

The magnitude is thus,

$\tau =(0.18: m) \cdot(85: N)\cdot\sin(65^{\circ})=14: N\cdot m$

The direction of torque is perpendicular to the plane of the radius and the force , and is given by Right Hand Rule by crossing the radius vector into the force vector. For this situation, the radius vector is left and downward and the force is downward, resulting in the direction of the torque out of the page.

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Question

A meter stick is nailed to a table at one end and is free to rotate in a horizontal plane parallel to the top of the table. Four forces of equal magnitude are applied to the meter stick at different locations. The figure below shows the view of the meter stick from above.

You may assume the forces and are applied at the center of the meter stick, and the forces and are applied at the end opposite the nail.

What is the relationship among the magnitudes of the torques on the meter stick caused by the four different forces?

Answer

Torque is given by,

$\tau =rFsin\theta$

Since all of the forces are equal in magnitude, the magnitude of the torque is then influenced by the radius r and the angle theta between the radius and the force.

For ,

$\tau_{1} =\left (\frac{1}{2}L \right )F\sin, 45^{\circ}=0.354\cdot L\cdot F$

For ,

$\tau_{2} =\left (\frac{1}{2}L \right )F\sin, 90^{\circ}=0.5\cdot L \cdot F$

For ,

$\tau_{3} =LF\sin, 30^{\circ}=0.5 \cdot L \cdot F$

For ,

$\tau_{4} =LF\sin, 90^{\circ}=L\cdot F$

Combining this information yields the relationship,

$\tau _{1} < \tau _{2} = \tau _{3} < \tau _{4}$

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Question

A man sits on the end of a long uniform metal beam of length . The man has a mass of and the beam has a mass of .

What is the magnitude of the net torque on the plank about the secured end of the beam? Use gravity .

Answer

The net torque on the beam is given by addition of the torques caused by the weight of the man and the weight of the beam itself, each at its respective distance from the end of the beam:

$\tau _{net}=\tau _{man}+\tau _{beam}$

Let's assign the direction of positive torque in the direction of the torques of the man's and the beam's weights, noting that they will add together since they both point in the same direction.

$\tau _{net}=L\cdot W_{man}\cdot\sin 90^{\circ}+\left ( \frac{1}{2} L\right )\cdot W_{beam}\cdot\sin, 90^{\circ}$

$\tau _{net}=L\cdot m_{man}\cdot g+\left ( \frac{1}{2} L\right )\cdot m_{beam}\cdot g$

We can further simplify by combining like terms:

$\tau _{net}=L\cdot g \cdot (m_{man}+\frac{1}{2} m_{beam})$

Using the given numerical values,

$\tau _{net}=(3.05: m)\cdot(9.8: m/s^{2})\cdot(85: kg+\frac{1}{2} 40: kg)=3140: N\cdot m$

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Question

Two children sit on the opposite sides of a seasaw at a playground, doing so in a way that causes the seasaw to balance perfectly horizontal. The child on the left is from the pivot.

What is the mass of the second child if she sits from the pivot?

Answer

A torque analysis is appropriate in this situation due to the inclusion of distances from a given pivot point. Generally,

$\tau_{net}=I\alpha$

This is a static situation. There are two torques about the pivot caused by the weights of two children. We will note that these weights cause torques in opposite directions about the pivot, such that

$\tau _{net}=0=\tau _{1}-\tau _{2}$

Consequently,

$\tau _{1}=\tau _{2}$

$r_{1}\cdot m_{1} \cdot g \cdot \sin: 90^{\circ}=r_{2}\cdot m_{2}\cdot g \cdot \sin: 90^{\circ}$

Or more simply,

$r_{1}\cdot m_{1}=r_{2} \cdot m_{2}$

Solving for ,

$m_{2}=\frac{m_{1}r_{1}}{r_{2}}=\frac{36: kg\cdot1.8: m}{2.4: m}}=27: kg$

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Question

A gymnast is practicing her balances on a long narrow plank supported at both ends. Her mass is . The long plank has a mass of .

Calculate the force the right support provides upward if she stands from the right end. Use gravity .

Answer

A torque analysis is appropriate in this situation due to the inclusion of distances from a given pivot point. Generally,

$\tau_{net}=I\alpha$

This is a static situation. As such, any pivot point can be chosen about which to do a torque analysis. The quickest way to the unknown force asked for in the question is to do the torque analysis about the left end of the plank. There are three torques about this pivot: two clockwise caused by the weights of the gymnast and the plank itself, and one counterclockwise caused by the force from the right support. Designating clockwise as positive,

$\tau _{net}=0=\tau _{gymnast}+\tau _{plank}-\tau _{right}$

Consequently,

$\tau _{gymnast}+\tau _{plank}=\tau _{right}$

$r _{gymnast}\cdot W_{gymnast}\cdot\sin: 90^{\circ}+r _{plank}\cdot W_{plank}\cdot \sin: 90^{\circ}=r _{right}\cdot F_{right}\cdot \sin: 90^{\circ}$

This simplifies to

$r _{gymnast}\cdot m_{gymnast}\cdot g+r _{plank}\cdot m_{plank}\cdot g=r _{right}\cdot F_{right}$

Solving for $F_{right}$ ,

$F_{right} = \frac{(r _{gymnast}\cdot m_{gymnast}+r _{plank}\cdot m_{plank})\cdot g}{r _{right}}$

Solving with numerical values,

$F_{right} = \frac{(2: m\cdot 55: kg+3: m\cdot 23: kg)\cdot 9.8: m/s^{2}}{6: m}=290: N$

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Question

A square of side lengths and mass is shown with possible axes of rotation.

Which statement of relationships among the moments of inertia is correct?

Answer

The moments of inertia for both axes and are equal because both of these axes are equivalently passing through the center of the mass.

By the parallel axis theorem for moments of inertia ( $I = I_{cm}+ MD^2$ ), the moment of inertia for axis is larger than or

because it is located a distance

$\frac{L}{2}$ away from the center of mass.

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Question

A wind catcher is created by attaching four plastic bowls of mass each to the ends of four lightweight rods, which are then secured to a central rod that is free to rotate in the wind. The four lightweight rods are of lengths , , , and .

Calculate the moment of inertia of the four bowls about the central rod. You may assume to bowls to be point masses.

Answer

The moment of inertia for a point mass is .

To calculate the total moment of inertia, we add the moment of inertia for each part of the object, such that

$I_{total}=I_{1}+I_{2}+I_{3}+I_{4}$

$I_{total}=m_{1}R_{1}^{2}+m_{2}R_{2}^{2}+m_{3}R_{3}^{2}+m_{4}R_{4}^{2}$

The masses of the bowls are all equal in this problem, so this simplifies to

$I_{total}=m (R_{1}^{2}+R_{2}^{2}+R_{3}^{2}+R_{4}^{2})$

Plugging in and solving with numerical values,

$I_{total}=(0.6: kg) [(0.8: m)^{2}+(0.6: m)^{2}+(0.4: m)^{2}+(0.2: m)^{2}]=0.72: kg\cdot m^{2}}$

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Question

A long uniform thin rod of length has a mass of .

Calculate the moment of rotational inertia about an axis perpendicular to its length passing through a point from one of its ends.

Answer

For a long thin rod about its center of mass,

$I_{cm}=\frac{1}{12}mL^{2}$

According to the parallel axis theorem,

$I_{//}=I_{cm}+md^{2}$

where is defined to be the distance between the center of mass of the object and the location of the axis parallel to one through the center of mass. For this problem, is the difference between the given distance and half the length of the rod.

Combining the above,

$I=\frac{1}{12}mL^{2}+m(\frac{1}{2}L-l)^{2}$

Inserting numerical values,

$I=\frac{1}{12}(12)(1.82)^{2}+(12)(\frac{1.82}{2}-.036)^{2}=6.9: kg\cdot m^{2}$

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