AP Physics B

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AP Physics 1 › AP Physics B

Questions 1 - 10

A man pulls a box up a incline to rest at a height of . He exerts a total of of work. What is the coefficient of friction on the incline?

We must know the mass of the box to solve

Explanation

Work is equal to the change in energy of the system. We are given the weight of the box and the vertical displacement, which will allow us to calculate the change in potential energy. This will be the total work required to move the box against gravity.

$W_g=\Delta PE=mg\Delta h$

The remaining work that the man exerts must have been used to counter the force of friction acting against his motion.

$W_{total}=W_g+W_f$

Now we know the work performed by friction. Using this value, we can work to solve for the force of friction and the coefficient of friction. First, we will need to use a second formula for work:

In this case, the distance will be the distance traveled along the surface of the incline. We can solve for this distance using trigonometry.

$\sin(\theta)=\frac{opp}{hyp}$

$\sin(30^o)=\frac{4m}{d}=$

$d=\frac{4m}{\frac{1}{2}}=8m$

We know the work done by friction and the distance traveled along the incline, allowing us to solve for the force of friction.

$F_f=\frac{70J}{8m}=8.75N$

Finally, use the formula for frictional force to solve for the coefficient of friction. Keep in mind that the force on the box due to gravity will be equal to $mg\cos(\theta)$ .

$F_f=\mu F_N=\mu(-F_g)$

$F_g=mg\cos(\theta)$

$F_f=\mu(-mg\cos(\theta))$

Plug in our final values and solve for the coefficient of friction.

$8.75N=-\mu(-50N)\cos(30^o)=\mu(50N)(0.866)$

$\mu=\frac{8.75N}{50N(0.866)}=0.20$

A ball is initially compressed against a spring $\left(k = 30 \frac{N}{m},\Delta x= 5cm\right)$ on a frictionless horizontal table. The ball is the released, and is shot to the right by the spring. Calculate the work done by the spring on the ball, .

Can not be determined.

Explanation

From the Work-energy theorem,

$W = \Delta KE$ .

From conservation of energy, we see that all of the potential energy contained in the spring will be transferred into kinetic energy of the ball, shown as

$\frac{1}{2}k \Delta x ^2= \frac{1}{2}m v^2 = KE_f$

Noting the initial kinetic energy of the ball is , we can see that the work done by the spring will be equal to the initial potential energy of the spring.

$W_s = \Delta KE = \frac{1}{2}mv^2= \frac{1}{2}k \Delta x ^2 = \frac{1}{2}30 \frac{N}{m}*(0.05m)^2= 37.5mJ$

Which of the following is not an acceptable unit for specific heat?

$\frac{J}{N\cdot K}$

$\frac{J}{g\cdot K}$

$\frac{cal}{g\cdot K}$

$\frac{Btu}{^oF\cdot lb}$

$\frac{J}{mol\cdot K}$

Explanation

Specific heat is most commonly applied in the equation to determine the heat required/released during a temperature change:

$Q=mc\Delta T$

Rearranging this equal, we can see that this units of specific heat are units of heat per units of mass times units of temperature.

$c=\frac{Q}{m\Delta T}$

Heat energy can have the units of Joules or calories. Mass can have units of grams or kilograms. Temperature can have units of Kelvin, degrees Celsius, or degrees Fahrenheit. British thermal units (Btu) is a less common unit of heat, defined by the amount of heat required to raise one pound of water by one degree Fahrenheit.

Newtons are a unit of force, and cannot be used in the units for specific heat.

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 $\mu$

$\mu=84$

Two asteroids in space are in close proximity to each other. Each has a mass of $6.69*10^{15}kg$ . If they are apart, what is the gravitational force between them?

$\small G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}$

$2.99*10^{11}N$

$3.01*10^{-8}N$

$5.98*10^{12}N$

$1.49*10^{-32}N$

$1.11*10^{30}N$

Explanation

To solve this problem, use Newton's law of universal gravitation:

$F_G=G\frac{m_1m_2}{r^2}$

We are given the constant, as well as the asteroid masses and distance (radius). Using these values we can solve for the force.

$F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{(6.69*10^{15}kg )(6.69*10^{15}kg )}{(100,000m)^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}(\frac{4.48*10^{31}kg^2}{10*10^9m^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}(4.48*10^{21}\frac{kg^2}{m^2})$

$F_G=2.99*10^{11}N$

Consider the following system:

Spinning rod with masses at end

Two spherical masses, A and B, are attached to the end of a rigid rod with length l. The rod is attached to a fixed point, p, which is at the midpoint between the masses and is at a height, h, above the ground. The rod spins around the fixed point in a vertical circle that is traced in grey. $\angle c$ is the angle at which the rod makes with the horizontal at any given time ( $c=0^{\circ}$ in the figure).

The rod is initially at rest in its horizontal position. How much work would it take to rotate the rod clockwise until it is vertical, at rest, and mass A is at the top?

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

Neglect air resistance and internal frictional forces. Ignore the mass of the rod itself.

None of these

Explanation

We can use the expression for conservation of energy:

Since the rod is both initially and finally at rest, we can removed both kinetic energies. Also, if we assume point p is at a height of 0, we can removed initial potential energy, leaving us with:

Plugging in the expression for potential energy and expanding for both masses:

Since the rod is vertical, we know that mass A is half a rod's length above our reference height, and mass B is half a rod's length below it. Thus we get:

$W = m_Ag\left (\frac{1}{2}l \right )+m_Bg\left (-\frac{1}{2}l \right )$

Factoring to clean up our expression:

$W = \frac{1}{2}gl\left ( m_A-m_B\right )$

We know all of our variables, so time to plug and chug:

$W = \frac{1}{2}(10\frac{m}{s^2})(4m)(10kg-6kg)$

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:

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 $\mu$

$\mu=84$

As a joke, Charlie glues C.J's phone to its receiver, which is bolted to her desk. Trying to extricate it, C.J. pulls on the phone with a force of for . She then pulls on the phone with a force of for . Unfortunately, all of her exertion is in vain, and neither the phone, nor receiver move at all. How much work did C.J. do on the phone in her 25 total seconds of pulling?

Explanation

Work is a measure of force and displacement $(W = F \cdot s)$ . Because C.J. did not move the phone at all, no work was done.

A semi-truck carrying a trailer has a total mass of . If it is traveling up a slope of $5^{\circ}$ to the horizontal at a constant rate of $20\frac{m}{s}$ , how much power is the truck exerting?

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

Explanation

Since the truck is traveling at a constant rate, we know that all of the power exerted by the truck is going into a gain in potential energy. The power exerted will be a function of the change in potential energy over time. Therefore, we can write the following formula:

$P = \frac{\Delta U}{t}=\frac{mg\Delta h}{t}$