The defintion of work is:

$\displaystyle W=F_{net}\cdot d$

The net force on this object is:

$\displaystyle F_{net}=F_{app}-F_{friction}$

We can calculate this term using the given values:

$\displaystyle F_{net}=53N-3.47N=49.53N$

The distance is given. Substituting these values:

$\displaystyle W=49.53 N (7.9m)=391.29 J$

The equation of work is given by $\displaystyle W = F\cdot d$, where $\displaystyle F$ is force and $\displaystyle d$ is the distance. Both the work and the force are given to us, but in vector form. In this case, we have to take the dot product of the force and distance.

When taking dot product, keep these rule in mind

$\displaystyle i \cdot i = 1, j \cdot j = 1, k \cdot k =1$

Every other dot proudct is equal to 0.

$\displaystyle (8i - 2j +3k) \cdot (2i + 3j - 1k) = \\((8\cdot2)+(-2\cdot3)+(3\cdot-1) ) = 7\:J$

Power is determined by calculating the work output per unit time. In this case, power will be the product of torque and angular velocity:

$\displaystyle P=\tau*\omega$

We are given values for our torque and our power output, allowing us to solve for the angualr velocity. Keep in mind that the power output is going to be 50% of the maximum.

$\displaystyle 250W=(30N\cdot m)(\omega)$

$\displaystyle \omega=8.33\frac{rad}{s}$

For this problem we can calculate the power as the product of force and velocity:

$\displaystyle P=Fv$

First, we need to find the velocity:

$\displaystyle v=\frac{\Delta x}{\Delta t}=\frac{20m}{10s}=2\frac{m}{s}$

Our force will be equal to the weight of the crate:

$\displaystyle F=mg=(50kg)(9.8\frac{m}{s})=490N$

Now, we can solve for power:

$\displaystyle P=Fv=(490N)(2\frac{m}{s})=980W$

Because the boxes are the same mass and are moving the same distance, the work done will remain the same between the two instances. Work does not depend on time:

$\displaystyle W=Fd$

However, when carrying the second box, the person moves more slowly. The overall time increases, which leads to a decrease in power.

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

The definition of power is:

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

The work done on the box to lift it is required in order to overcome the force of gravity. If we can find the force of gravity on the object, we can calculate the net force. The force of gravity on any object near the Earth's surface is:

$\displaystyle F_g=mg$

The definition of work is:

$\displaystyle W=F_{net}d$

We can substitute the force of gravity for the net force, resulting in the equation:

$\displaystyle W=mgd$

Substituting this into our power equation, we find:

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

Plugging in our given values and constants, we find:

$\displaystyle P=\frac{(25 kg) (9.81 \frac{m}{s^2})(5.76m)}{15 s}$

$\displaystyle P=\frac{(25)(9.81)(5.76)}{15}\frac{kgm^2}{s^3}$

$\displaystyle P=87.36 W$

The conservation of energy equation is $\displaystyle K_1+U_1=K_2+U_2$.

The ball starts from rest so $\displaystyle K_1=0$. It starts at a height of 150m, so $\displaystyle U_1=mgh$.  When the ball reaches the bottom, height is zero and thus, $\displaystyle U_2=0$ and $\displaystyle K_2=\frac{1}{2}mv^2$.  The conservation of energy equation can be adjusted below.

$\displaystyle U_1=K_2$

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

Solve for v.

$\displaystyle v=\sqrt{2gh}$

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

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

Conservation of energy states that $\displaystyle K_1+U_1=K_2+U_2$.

The skateboarder starts from rest; thus, $\displaystyle K_1=0$ and $\displaystyle U_1=mgh$. At the bottom of the incline, $\displaystyle K_2=\frac{1}{2}mv^2$ and $\displaystyle U_2=0$.

$\displaystyle U_1=K_2$

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

Solve for v.

$\displaystyle v=\sqrt{2gh}$

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

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

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

Relevant equations:

$\displaystyle K_i+U_i=K_f+U_f$

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

$\displaystyle U = mgh$

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

$\displaystyle K_i = 0$

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

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

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

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

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

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

Solve for the final velocity.

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

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

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

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

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

It is given to us that the work done by the drag force is $\displaystyle -3800\:J$ which means that work is done in the opposite direction. We take the net work by adding the two works together we get, $\displaystyle 1100\:J$ 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:

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

$\displaystyle m$ is the mass of the block and we are trying to solve for $\displaystyle v$.

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