AP Physics 1

A rocket of mass is motionless at the origin. The rocket then fires. At location , it is traveling at $<250,250>\frac{m}{s}$ .

Determine the magnitude of the average force of the rocket.

None of these

Explanation

Determining magnitude of final velocity:

$|\overrightarrow{v}|=\sqrt{v_x^2+v_y^2}$

Plugging in values:

$|\overrightarrow{v}|=\sqrt{250^2+250^2}$

$|\overrightarrow{v}|=353.6\frac{m}{s}$

Using definition of kinetic energy:

Plugging in values:

Since there was no initial kinetic energy:

Finding total distance:

$|\overrightarrow{d}|=\sqrt{d_x^2+d_y^2}$

$|\overrightarrow{d}|=\sqrt{500^2+500^2}$

Using

Plugging in values:

How much work (in kilojoules) is done to accelerate a car (3000kg) from rest to $60\frac{m}{s}$ .

5400 kilojoules

3000 kilojoules

6200 kilojoules

3400 kilojoules

2300 kilojoules

Explanation

Work is found by finding the change in kinetic energy. Since the car started from rest it had no initial kinetic energy.

$KE=\Delta.5(3000kg)(60)^2=5400000J$

Divide by 1000 to convert to kilojoules and get 5400 kilojoules.

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

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

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

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

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

$W_{outside}=-1710J$

Consider the following system:

Slope_2

If the mass is and $A = 50^{\circ}$ , what is the tension, ? Assume no frictional forces.

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

Explanation

Since there is no friction between the mass and slope, there are only two relevant forces acting on the mass: gravity and tension. Furthermore, since the block is not in motion, we know that these forces are equal to each other. Therefore:

Substituting in an expression for the force of gravity, we get:

$T = mgsin(\theta)$

We know all of these values, allowing us to solve for the tension:

$T = (5kg)(10\frac{m}{s^2})sin(50^{\circ})$

A woman has a jet pack that has a mass of . She ignites the jet pack and it accelerates her at a rate of $5\frac{m}{s^2}$ . Determine the total work done by the jet pack when she has reached a height of .

None of these

Explanation

$F_{net}=ma=F_{jetpack}+F_{gravity}$

$ma=F_{jetpack}+mg$

$(55+5)*5=F_{jetpack}+(55+5)*-9.8$

$F_{jetpack}=888N$

$W_{jetpack}=F_{jetpack}*d$

$W_{jetpack}=888*10$

$W_{jetpack}=8880J$

A man throws a pizza in the air. If he released it at a height of , and his throwing motion was a distance of directly up, determine the average force of the man on the pizza during the throw.

None of these

Explanation

Based on the information given, his throw would have started at above the ground. The pizza gained a maximum height of . Thus, the gravitational potential energy in relation to the starting position will be:

$PE_{gravitational}=mg(h_f-h_i)$

$PE_{gravitational}=.55*9.8(3-.5)$

$PE_{gravitational}=13.5J$

Since at it's maximum height, all of the energy will be gravitational, the work done on the pizza will be equal to the potential energy, thus:

Using

A rocket of mass is motionless at the origin. The rocket then fires. At location , it is traveling at $<250,250>\frac{m}{s}$ .

Determine the work done on the rocket

None of these

Explanation

Determining magnitude of final velocity:

$|\overrightarrow{v}|=\sqrt{v_x^2+v_y^2}$

Plugging in values:

$|\overrightarrow{v}|=\sqrt{250^2+250^2}$

$|\overrightarrow{v}|=353.6\frac{m}{s}$

Using definition of kinetic energy:

Plugging in values:

Since there was no initial kinetic energy:

A rocket is in space at location when it fires it's thrusters. The thrusters provide a force of . The thusters are turned off at location .

What is the work done on the rocket by the thrusters?

$5.1\times10^5 J$

None of these

Explanation

Because all of the force was in the Y direction, all of the work will come from the change in the Y coordinate.

We will use the definition of work

$W=F\cdot d$

First we need to find the distance traveled.

Then we can plug that into our work equation.

$52 m\cdot 9900 N = 5.1\times10^5 J$

A rocket is in space at location when it fires it's thrusters. The thrusters provide a force of . The thusters are turned off at location .

What is the work done on the rocket by the thrusters?

$5.1\times10^5 J$

None of these

Explanation

Because all of the force was in the Y direction, all of the work will come from the change in the Y coordinate.

We will use the definition of work

$W=F\cdot d$

First we need to find the distance traveled.

Then we can plug that into our work equation.

$52 m\cdot 9900 N = 5.1\times10^5 J$

If three locomotives are pulling a train, how much power does each locomotive need to apply on average during the first second to accelerate the train at $1\frac{cm}{s^2}$ from rest?