AP Physics 1 - Work | Practice Hub

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 $v_{initial}=50 \frac{m}{s}$

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

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

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

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

$W_{outside}=-1710J$

During takeoff, a rocket goes from $0\frac{m}{s}$ at of elevation to $2{\frac{m}{s}}$ at of elevation.

Estimate the amount of work done by the rocket engine. Neglect any mass change due to burning of fuel.

Explanation

Initially there is only potential energy, and in the final state there is both kinetic and potential.

Plug in all values and solve:

A train of mass $5.11*10^{11}kg$ goes from $50 \frac{km}{hr}$ to $0 \frac{km}{hr}$ in . Calculate the magnitude of force from the brakes.

$4.93*10^{10}$

$9.15*10^{10}$

$5.55*10^{10}$

$7.86*10^{10}$

$2.18*10^{10}$

Explanation

Use work:

All energy will be kinetic.

Convert $\frac{km}{hr}$ to $\frac{m}{s}$ :

$\frac{50km}{hr}*\frac{1000m}{1km}*\frac{1 hr}{3600 s}=13.9\frac{km}{hr}$

$\frac{0km}{hr}*\frac{1000m}{1km}*\frac{1 hr}{3600 s}=0\frac{km}{hr}$

Plug in values. Force will be negative as it is directed against the direction of travel:

$F*1000=.5*5.11*10^{11}*0^2-.5*5.11*10^{11}*(13.9)^2$

Solve for :

$F=-4.93*10^{10}$

$|F|=4.93*10^{10}$

A car of mass is accelerated from $10 \frac{m}{s}$ to $35 \frac{m}{s}$ in 2s.

Determine the work done on the car in this time frame.

None of these

Explanation

Use the definition of work:

$W=\Delta KE$

Plug in known values and solve.

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.

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

A large block is sitting on the floor. The block is . You pull the block with a rope, applying of force at a 25 degree angle with respect to the horizon. How much work did you do on the block if you moved it ? The block moves at a constant speed while you are pulling it.

Explanation

First let's draw a free body diagram of the block with all the forces acting on it.

Fbd

The block has a weight force which is . The normal force is equal and opposite (otherwise the block would be accelerating into the ground or into the air. The applied force is at an angle, and in order to find the work done we need to find the applied force that is pulling the block across the floor. This is:

$F_{x}=F_{applied} cos(\theta)=45.32N$

The work done is just the product of the applied force and the distance the block slides,

$W=F\cdot \Delta x=45.32N \cdot 5m=226.6J$

Work is an energy, and the units of $N\cdot m$ are equal to Joules.

A 50kg man pushes against a wall with a force of 100N for 10 seconds. How much work does the man accomplish?

Explanation

The answer is because no work is done. For work to be done a force must be exerted across a distance parallel to the direction of the force. In this case, a force is exerted by the man but the wall is stationary and since it does not move there is no distance for work to take place on. The formula for work can be written as:

$W=f sin{\theta}d$

Here, $\theta$ is zero, so $sin{\theta}$ is also zero, making the entire term, and thus the work zero.

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.

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

Juri is tugging her wagon behind her on the way to... wherever her wagon needs to go. The wagon repair shop. She has a trek ahead of her--five kilometers--and she's pulling with a force of 200 newtons. If she's pulling at an angle of 35 degrees to the horizontal, what work will be exerted on the wagon to get to the repair shop?

Explanation

Work exerted on an object is equal to the dot product of the force and displacement vectors, or the product of the magnitudes of the vectors and the sin of the angle between them:

$W=F\cdot d=Fdsin(\theta)$

The work exerted on the wagon in this problem is thus:

$W=200N(5km)sin(35^{\circ})$

Raul is pushing a broken down car across the flat expanse of the Mojave to his shop.

If his shop is three kilometers away and he pushes with a Herculean force of one thousand newtons in the direction of his shop, how much work will be done on the car?

Explanation

Work is given by the dot product of force and displacement. Since both the force and displacement are in the same direction in this problem, work is simply the product of the two:

$Work=3\ km(1000N)$

$Work=3000\ kJ$

Work

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