Question

1. a 45.0 - kilogram block rests motionless on a horizontal, frictionless table. how much work must be done to accelerate the block to a speed of 65.00 meters per second?

1. if box a has a mass of m and rests a height of 2h off the ground. box b has a mass of 2m and rests a height of h off the ground. what is the relationship between the boxes potential energies? (hint: what is the formula for potential energy? use the formula to help! draw a picture if you need to)

1. a force of 89 newtons is required to move an object at a constant speed of 25 meters per second. what is the power supplied, in watts?

Explanation:

Question 4

Step1: Recall Work - Energy Theorem

The work - energy theorem states that the work done on an object is equal to the change in its kinetic energy. The formula for kinetic energy is $KE=\frac{1}{2}mv^{2}$, where $m$ is the mass of the object and $v$ is its velocity. Initially, the block is motionless, so its initial kinetic energy $KE_{i} = 0$. The final kinetic energy $KE_{f}=\frac{1}{2}mv^{2}$. The work done $W$ is equal to the change in kinetic energy, so $W=\Delta KE=KE_{f}-KE_{i}=\frac{1}{2}mv^{2}-0=\frac{1}{2}mv^{2}$.

Step2: Substitute the values

We are given that $m = 45.0\space kg$ and $v=65.00\space m/s$. Substitute these values into the formula for work:

$$
LATEXBLOCK0
$$

Step1: Recall the formula for gravitational potential energy

The formula for gravitational potential energy (assuming near - Earth's surface) is $PE = mgh$, where $m$ is the mass of the object, $g$ is the acceleration due to gravity, and $h$ is the height of the object above the reference level (in this case, the ground).

Step2: Calculate the potential energy of Box A and Box B

For Box A: $m_{A}=m$, $h_{A} = 2h$. So $PE_{A}=m_{A}gh_{A}=m\times g\times2h = 2mgh$.

For Box B: $m_{B}=2m$, $h_{B}=h$. So $PE_{B}=m_{B}gh_{B}=2m\times g\times h=2mgh$.

Step3: Compare the potential energies

Since $PE_{A}=2mgh$ and $PE_{B} = 2mgh$, we can see that $PE_{A}=PE_{B}$.

Step1: Recall the formula for power

The formula for power $P$ in terms of force $F$ and velocity $v$ (when the force is in the direction of motion and the speed is constant) is $P = Fv$. This is because power is defined as the rate of doing work, and $W = Fd$ (work is force times distance), and $P=\frac{W}{t}=\frac{Fd}{t}=F\times\frac{d}{t}=Fv$ (since $v = \frac{d}{t}$).

Step2: Substitute the values

We are given that $F = 89\space N$ and $v = 25\space m/s$. Substitute these values into the formula for power:

$$
LATEXBLOCK0
$$

Answer:

The work done is $95062.5$ joules.

Question 5