Kinetic Energy

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AP Physics C: Electricity and Magnetism › Kinetic Energy

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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?

CORRECT

Explanation

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:

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

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?

CORRECT

Explanation

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.

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

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.

$54.25 \frac{m}{s}$

CORRECT

$55.0 \frac{m}{s^2}$

$51.65 \frac{m}{s}$

$51.65 \frac{m}{s^2}$

$0 \frac{m}{s}$

Explanation

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:

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