Energy - AP Physics C Electricity & Magnetism

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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

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

Calculate how much potential energy a vertical standing spring gains if someone puts a 4kg box on top of it. Take the spring constant to be 120N/s.

Answer

First, find out how much the spring compresses when the 4kg box is put on top of it. To find this, consider what forces are acting on the box when it is resting on the spring. The forces acting on it are the upward spring force, , and the downward gravitational force,; thus, the net force acting on the box is given by the equation below.

$F_{net}=F_s-F_g=-kx-mg=0$

We have chosen the upward direction to be positive and the downward direction to be negative. The net force is equal to zero because the box is at rest. Solve for x.

$x=-\frac{mg}{k}$

$x=-\frac{(4\ kg)(-9.8\ \frac{m}{s^2})}{120\ \frac{N}{m}}$

$x=0.33\ \text{m}$

Then use this value to find the potential energy of the spring.

$U=\frac{1}{2}kx^2$

$U=\frac{1}{2}(120\ \frac{N}{m})(0.33\ m)^2$

$U=6.5\ J$

So when the box is placed on top of the spring, the spring gains a potential energy of 6.5J.

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Question

A man throws a 5kg ball straight up. At the top of its trajectory, the ball has a potential energy of 160 Joules. At what velocity does the man initially throw the ball?

Answer

We can use conservation of energy to find the velocity required to launch the ball.

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

$PE = KE \rightarrow 160 = \frac{1}{2}mv^2$

$v = 8\ \frac{m}{s}$

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Question

A man throws a 5kg ball straight up. At the top of its trajectory, the ball has a potential energy of 160 Joules. What is the highest point the ball reaches if it is thrown from the level of the ground?

$g=10\frac{m}{s^2}$

Answer

The equation for potential energy is:

We can rearrange to solve for the height:

$h=\frac{PE}{mg}$

Plug in our given values to solve:

$h=\frac{160J}{(5kg)(10\frac{m}{s^2})}=3.2m$

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Question

$G \approx (20/3) 10^{-11} \ \frac{m^3}{kgs^2}, \quad M_{earth} \approx 610^{24}\ kg, \quad R_{earth} \approx 6000 \ km$

A man launches a 5kg ball straight up from sea level. Disregarding drag, if the man would like to launch the ball into an escape trajectory from earth's gravity, approximately how fast must he launch the ball (in meters per second)?

$G=6.6710^{-11}\frac{m^3}{s^2kg}$

$M_E=6*10^{24}kg$

Answer

In order to launch the ball at a velocity that escapes Earth's gravity, the kinetic energy must overcome the gravitational pull of the Earth. We can use conservation of energy to make the kinetic energy and gravitational potential energies equivalent.

$KE_{escape} = \frac{1}{2}m v_{escape}^2$

$U_{gravitational} = \frac{GMm}{r}$

$KE_{e} = U_{g}$

$\frac{1}{2}mv_{e}^2= \frac{GMm}{r}$

Rearrange to solve for the escape velocity:

$v_e = \sqrt{\frac{2GM}{r}}$

Finally, plug in our values and solve.

$v_e= \sqrt{\frac{2(6.6710^{-11})(610^{24})}{610^6}}$

$v_e=\sqrt{\frac{72}{6}10^{7}} =\sqrt{1.210^{8}} \approx\sqrt{12110^{6}} = 11000\ \frac{m}{s}$

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Question

Calculate how much potential energy a vertical standing spring gains if someone puts a 4kg box on top of it. Take the spring constant to be 120N/s.

Answer

First, find out how much the spring compresses when the 4kg box is put on top of it. To find this, consider what forces are acting on the box when it is resting on the spring. The forces acting on it are the upward spring force, , and the downward gravitational force,; thus, the net force acting on the box is given by the equation below.

$F_{net}=F_s-F_g=-kx-mg=0$

We have chosen the upward direction to be positive and the downward direction to be negative. The net force is equal to zero because the box is at rest. Solve for x.

$x=-\frac{mg}{k}$

$x=-\frac{(4\ kg)(-9.8\ \frac{m}{s^2})}{120\ \frac{N}{m}}$

$x=0.33\ \text{m}$

Then use this value to find the potential energy of the spring.

$U=\frac{1}{2}kx^2$

$U=\frac{1}{2}(120\ \frac{N}{m})(0.33\ m)^2$

$U=6.5\ J$

So when the box is placed on top of the spring, the spring gains a potential energy of 6.5J.

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Question

A man throws a 5kg ball straight up. At the top of its trajectory, the ball has a potential energy of 160 Joules. At what velocity does the man initially throw the ball?

Answer

We can use conservation of energy to find the velocity required to launch the ball.

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

$PE = KE \rightarrow 160 = \frac{1}{2}mv^2$

$v = 8\ \frac{m}{s}$

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Question

A man throws a 5kg ball straight up. At the top of its trajectory, the ball has a potential energy of 160 Joules. What is the highest point the ball reaches if it is thrown from the level of the ground?

$g=10\frac{m}{s^2}$

Answer

The equation for potential energy is:

We can rearrange to solve for the height:

$h=\frac{PE}{mg}$

Plug in our given values to solve:

$h=\frac{160J}{(5kg)(10\frac{m}{s^2})}=3.2m$

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Question

$G \approx (20/3) 10^{-11} \ \frac{m^3}{kgs^2}, \quad M_{earth} \approx 610^{24}\ kg, \quad R_{earth} \approx 6000 \ km$

A man launches a 5kg ball straight up from sea level. Disregarding drag, if the man would like to launch the ball into an escape trajectory from earth's gravity, approximately how fast must he launch the ball (in meters per second)?

$G=6.6710^{-11}\frac{m^3}{s^2kg}$

$M_E=6*10^{24}kg$

Answer

In order to launch the ball at a velocity that escapes Earth's gravity, the kinetic energy must overcome the gravitational pull of the Earth. We can use conservation of energy to make the kinetic energy and gravitational potential energies equivalent.

$KE_{escape} = \frac{1}{2}m v_{escape}^2$

$U_{gravitational} = \frac{GMm}{r}$

$KE_{e} = U_{g}$

$\frac{1}{2}mv_{e}^2= \frac{GMm}{r}$

Rearrange to solve for the escape velocity:

$v_e = \sqrt{\frac{2GM}{r}}$

Finally, plug in our values and solve.

$v_e= \sqrt{\frac{2(6.6710^{-11})(610^{24})}{610^6}}$

$v_e=\sqrt{\frac{72}{6}10^{7}} =\sqrt{1.210^{8}} \approx\sqrt{12110^{6}} = 11000\ \frac{m}{s}$

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Question

A 0.8kg ball is dropped from rest from a cliff that is 150m high. Use conservation of energy to find the vertical velocity of the ball right before it hits the bottom of the cliff.

Answer

The conservation of energy equation is .

The ball starts from rest so . It starts at a height of 150m, so . When the ball reaches the bottom, height is zero and thus, and $K_2=\frac{1}{2}mv^2$ . The conservation of energy equation can be adjusted below.

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

Solve for v.

$v=\sqrt{2gh}$

$v=\sqrt{2(9.8\ \frac{m}{s^2})(150\ m)}$

$v=54.2\ \frac{m}{s}$

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Question

Starting from rest, a skateboarder travels down a 25o incline that's 22m long. Using conservation of energy, calculate the skateboarder's speed when he reaches the bottom. Ignore friction.

Answer

Conservation of energy states that .

The skateboarder starts from rest; thus, and . At the bottom of the incline, $K_2=\frac{1}{2}mv^2$ and .

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

Solve for v.

$v=\sqrt{2gh}$

Using trigonometry, $h=22\sin(25)$ .

$v=\sqrt{2(9.8\ \frac{m}{s^2})(22)(\sin(25))}$

$v=13.5\ \frac{m}{s}$

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Question

A $\small 5.0kg$ bowling ball is dropped from $\small 5.0m$ above the ground. What will its velocity be when it is $\small 1.0m$ above the ground?

Answer

Relevant equations:

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

Determine initial kinetic and potential energies when the ball is dropped.

$U_i = (5kg)(9.8\frac{m}{s^2})(5m)=245J$

Determine final kinetic and potential energies, when the ball has fallen to $\small 1.0m$ above the ground.

$K_f=\frac{1}{2}(5kg)v_f^2$

$U_f = (5kg)(9.8\frac{m}{s^2})(1m)=49J$

Use conservation of energy to equate initial and final energy sums.

$0 + 245J = \frac{1}{2}(5kg)v_f^2+49J$

Solve for the final velocity.

$0 + 245J = \frac{1}{2}(5kg)v_f^2+49J$

$v_f^2=\frac{245J-49J}{\frac{1}{2}(5kg)}=78.4\frac{m^2}{s^2}$

$v_f=8.85 \frac{m}{s} \approx 8.9\frac{m}{s}$

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Question

A solid metal object with mass of is dropped from rest at the surface of a lake that is deep. The water exerts a drag force on the object as it sinks. If the total work done by the drag force is -, what is the speed of the object when it hits the sand at the bottom of the lake?

Answer

This is a conservation of energy problem. First we have to find the work done by gravity. This can be found using:

$W= F\cdot d, F = mg \ W = 25:kg \cdot 9.8:m/s^2 \cdot 20:m = 4900:J$

It is given to us that the work done by the drag force is which means that work is done in the opposite direction. We take the net work by adding the two works together we get, of net work done on the block.

Since this is a conservation of energy problem, we set the net work equal to the kinetic energy equation:

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

is the mass of the block and we are trying to solve for .

$v = \sqrt{\frac{2 \cdot 1100}{25}} = 9.38:m/s$

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