AP Physics 1 - Newtonian Mechanics

Determine the work done by nonconservative forces if an object with mass 10kg is shot up in the air at $30\frac{m}{s}$ returns to the same height with speed $27\frac{m}{s}$ .

Explanation

Since this question refers to work done by nonconservative forces, we know that:

$W_{outside}=\Delta PE+\Delta KE$

Here, $\Delta PE$ is the change in potential energy, and $\Delta KE$ is a change in kinetic energy. $\Delta PE$ is because the object returns to the same height as when it was launched. $\Delta KE$ however has changed because the object's velocity has changed. Recall that the formula for the change in kinetic energy is given by:

$\Delta KE= mv^2_{final}-mv^2_{initial}$

Here is the mass of the object, $v_{final}$ is the final velocity of the object and $v_{initial}$ is the initial velocity of the object.

In our case:

, $v_{final}=45 \frac{m}{s}$ , and $v_{initial}=50 \frac{m}{s}$

$\Delta KE= mv^2_{final}-mv^2_{initial}$

$\Delta KE=10kg*\left(27\frac{m}{s}\right)^2-10kg*\left(30\frac{m}{s}\right)^2$

$\Delta KE= 7290J-9000J= -1710J$

$W_{outside}=\Delta PE+\Delta KE$

$W_{outside}=-1710J$

A model rocket weighing 5kg has a net propulsion force of 50N. Over a small period of time, the rocket speeds up (with constant acceleration) from an initial velocity of $20\frac{m}{s}$ to a final velocity of $25\frac{m}{s}$ . Let us assume that the loss of mass due to fuel consumption is negligible and that the net force is along the direction of motion. How much net work was done on the rocket?

Explanation

The net work done is equal to the change in kinetic energy. So we must find the kinetic energy at both the initial and final velocities and subtract.

$W=\Delta KE=1562.5-1000={562.5J}$

An 500kg elevator is at rest. If it is raised 50 meters and returns to rest, how much total work was done on the elevator?

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

Explanation

This can be a tricky question. You need to rely on the work-energy theorem, which states:

$W_{net}= \Delta K$

Since the elevator is at rest at both the beginning and end, the net work is 0; there is no net change in energy, and therefore no work.

This theorem can be confusing to some since it completely negates potential energy. However, let's think about the situation presented in the problem. A force is required to raise the elevator, meaning that energy is put into the system. However, since it comes back to rest, all of the energy that was put in has been removed by the force of gravity, resulting in a net of zero work.

A ball of mass is thrown at a target. The ball strikes with a velocity of $20\frac{m}{s}$ and bounces back with equal magnitude. Determine the work done on the ball.

None of these

Explanation

There is no net work done on the ball. The wall stopped the ball, doing negative work, then accelerated the ball, doing positive work. These end up canceling each other out.

Consider a constant force, given below, that acts on an object as it moves along the path $\vec{r}$ , also given below. Calculate the work done on the object. The units of the force and path are and , respectively.

$\vec{F}=20\hat{i}-3\hat{j}(N)$

$\vec{r}=4\hat{i}+10\hat{j}(m)$

Explanation

In order to find the work done on the object, we need to take the dot product of the force and the path taken. This is a direct calculation.

$W=\vec{F}\cdot \vec{r}=\left( 20\hat{i}-3\hat{j}\right) \cdot \left( 4\hat{i}+10\hat{j}\right )$

Rollercoaster

The height at the top of the hill is

If the velocity of the train at the top of the loop is $28\frac{m}{s }$ , how high is the highest point of the loop?

Explanation

Due to conservation of energy, all energy is conserved throughout this problem. When the train is at the top of the hill, all energy is stored in the form of gravitational potential energy, represented by the equation:

$PE_{_{grav}}=mgh$

Once the train starts rolling down the hill some of that energy gets transferred to kinetic energy, which is represented by the equation:

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

Because you are given the velocity at the top of the loop it is possible to find the height of the loop by connecting the two equations as follows:

$mgh_{0}=\frac{1}{2}mv^{2}+mgh_{f}$

Where $h_{0}$ is the height at the top of the hill, and $h_{f}$ is the height at any given point in the system. By plugging in the given values and cancelling out the for mass on both sides of the equation, it is possible to find the height at the top of the loop as follows:

$9.81*60=\frac{1}{2}(28^{2})+9.81h_{f}$

$h_{f}\approx20m$

Mass of Mars: $6.39*10^{23}kg$

Universal gravitation constant: $6.67*10^{-11}\frac{N\cdot m^2}{kg^2}$

Radius of Mars:

A buggy on the surface of Mars locks up it's break and slides on Martian ice. If the buggy was traveling at $25\frac{m}{s}$ and took to stop. Determine the coefficient of friction between Martian ice and the tires.

None of these

Explanation

The normal force will be equal to the magnitude of the force of gravity pushing the buggy into the ground.

$F_N=|-G\frac{m_1m_2}{r^2}|$

Using definition of frictional force and work equation:

$.5mv_i^2-F_N*d*\mu=.5mv^2_f$

Combining equations:

$.5mv_i^2-G\frac{m_1m_2}{r^2}*d*\mu=.5mv^2_f$

Canceling out the mass of the buggy and plugging in values:

$.5(25)^2-6.67*10^{-11}\frac{6.39*10^{23}}{(3.39*10^6)^2}*250*\mu=0$

Solving for $\mu$

$\mu=84$

A blue rubber ball weighing is rolling with a velocity of $2\frac{m}{s}$ when it hits a still red rubber ball with a weight of . After this elastic collision, what are the speeds and directions of the blue and red balls respectively?

$\frac{2}{3} \frac{m}{s}$ to the left, $\frac{4}{3} \frac{m}{s}$ to the right

$\frac{4}{3} \frac{m}{s}$ to the left, $\frac{2}{3} \frac{m}{s}$ to the right

$\frac{4}{3} \frac{m}{s}$ to the right, $\frac{2}{3} \frac{m}{s}$ to the left

At rest, $2 \frac{m}{s}$ to the right

At rest, $1 \frac{m}{s}$ to the right

Explanation

Because we are solving for two velocities (two unknowns), we need two equations. We can use the conservation of linear momentum:

$m_{1}v_{1}+m_{2}v_{2} = m_{1}v_{1}'+m_{2}v_{2}'$

Because we know that the collision is elastic, we know that kinetic energy is conserved:

$\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}=\frac{1}{2}m_{1}v_{1}'^{2}+\frac{1}{2}m_{2}v_{2}'^{2}$

The red ball starts at rest so

$v_{2}=0$

The above equations can then be simplified and one can solve for $v_{1}'$ and $v_{2}'$ .

$v_{1}'=\left ( \frac{m_{1}-m_{2}}{m_{1}+m_{2}} \right )v_{1}=\left ( \frac{20-40}{20+40} \right )2=-\frac{2}{3}$

Negative implies the ball is moving to the left.

$v_{2}'=\left ( \frac{2m_{1}}{m_{1}+m_{2}} \right )v_{1}=\left ( \frac{2(20)}{20+40} \right )2=\frac{4}{3}$

A spaceship of mass is motionless in space. The rocket is turned on and provides a constant force of . Assume the mass of mass due to spent fuel is negligible.

Determine the distance traveled in .

None of these

Explanation

Plugging in values

$a=.133\frac{m}{s}$

Using

$v_f=1.33\frac{m}{s}$

Using

Initially the space ship is motionless, so it has kinetic energy

$W=6633N\cdot m$

$W=F*d=6633N\cdot m$

$W=1000*d=6633N\cdot m$

A spaceship of mass is motionless in space. The rocket is turned on and provides a constant force of . Assume the mass of mass due to spent fuel is negligible.

Determine the distance traveled in .

None of these

Explanation

Plugging in values

$a=.133\frac{m}{s}$

Using

$v_f=1.33\frac{m}{s}$

Using

Initially the space ship is motionless, so it has kinetic energy

$W=6633N\cdot m$

$W=F*d=6633N\cdot m$

$W=1000*d=6633N\cdot m$

Newtonian Mechanics

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