First, we calculate the initial kinetic energy of the car:

$\displaystyle KE_{initial}=\frac{1}{2}mv_{initial}^2$

$\displaystyle KE_{initial}=\frac{1}{2}(1000kg)(20\frac{m}{s})^2=200,000J$

Then, calculate the final kinetic energy after the car has reached a velocity of $\displaystyle 35\frac{m}{s}$:

$\displaystyle KE_{final}=\frac{1}{2}mv_{final}^2$

$\displaystyle KE_{final}=\frac{1}{2}(1000kg)(35\frac{m}{s})^2=612,500J$

To find the change in kinetic energy, we subtract the initial energy from the final energy.

$\displaystyle 612,500J - 200,000J = 412,500J$

This answer in positive because the car gains energy when it increases its velocity.

The first thing we need to find is the final velocity of the car. We know it starts from rest and has an acceleration of $\displaystyle 10\frac{m}{s^2}$ for five seconds. We can calculate the final velocity using these values and the appropriate kinematics equation.

$\displaystyle v_f = v_i + at$

$\displaystyle v_f=0\frac{m}{s}+(10\frac{m}{s^2})(5s)=50\frac{m}{s}$

The formula for kinetic energy is:

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

We can use the calculated final velocity and the mass of the car to determine its final kinetic energy.

$\displaystyle (\frac{1}{2})(1500kg)(50\frac{m}{2})^2 = 1,875,000J$

The total kinetic energy for an object rolling without slipping is the sum of its translational and rotational kinetic energies.

The translational kinetic energy treats the object as if it were a mass sliding along at a constant velocity. We can calculate the translational kinetic energy using the equation:

$\displaystyle KE_t=\frac{1}{2}mv^2$

Use this equation with the given mass and velocity to solve for translational kinetic energy.

$\displaystyle KE_t=\frac{1}{2}(5kg)(10\frac{m}{s})^2$

$\displaystyle KE_t=250J$

The rotational kinetic energy is given by the formula:

$\displaystyle KE_r=\frac{1}{2}I\omega ^2$

In this equation, $\displaystyle \small I$ is the moment of inertia and $\displaystyle \small \omega$ is the angular velocity of the rolling object in radians per second (or Hz, since radians are a dimensionless unit). For the uniform thin hoop, the rolling moment of inertia around a perpendicular axis through the center of the hoop is given by the equation:

$\displaystyle I=MR^2$

The relationship between angular and linear velocity in the case of a hoop rolling without slipping is:

$\displaystyle \omega=\frac{v}{R}$

Solve for the moment of inertia and the angular velocity of the hoop.

$\displaystyle I=MR^2=(5kg)(0.5m)^2=1.25kg*m^2$

$\displaystyle \omega=\frac{v}{R}=\frac{10\frac{m}{s}}{0.5m}=20Hz$

Use these values in the formula for rotational kinetic energy.

$\displaystyle KE_r=\frac{1}{2}I\omega ^2$

$\displaystyle KE_r=\frac{1}{2}(1.25kg*m^2)(20Hz) ^2$

$\displaystyle KE_r=250J$

The total kinetic energy is the sum of the translational and rotational kinetic energies.

$\displaystyle KE=KE_t+KE_r$

$\displaystyle KE=250J+250J=500J$

To find average power, we find the total work done by the crane in lifting the steel beam, and then divide it by the time taken to complete the task.

$\displaystyle P=\frac{W}{t}$

In this case, the total work done is equal to the change in potential energy of the steel beam. for this, we use the potential energy equation:

$\displaystyle W=\Delta PE=mg\Delta h$

$\displaystyle W=(500kg)(9.8\frac{m}{s^2})(20m)$

$\displaystyle W=98,000J$

We then divide the work by $\displaystyle 10 s$ to find average power:

$\displaystyle P_{avg}=\frac{98,000J}{10s}$

$\displaystyle P_{ave}=9,800\frac{J}{s}=9,800W$

First we find the total work required to lift the crate:

$\displaystyle W=\Delta PE=mg\Delta h$

We are given the mass, change in height, and the acceleration of gravity. Using these values, we can solve for the total work done.

$\displaystyle W=(100kg)(9.8\frac{m}{s^2})(10m)=9800J$

Next, we can use the equation for power to find the time to complete this task if the engine is working at its maximum power.

$\displaystyle P=\frac{\Delta W}{\Delta t}$

$\displaystyle 2000W=\frac{9800J}{\Delta t}$

$\displaystyle \Delta t=\frac{9800J}{2000W}=4.9s$