AP Physics 1 : Specific Forces

Study concepts, example questions & explanations for AP Physics 1

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Example Questions

Example Question #676 : Newtonian Mechanics

Given that the coefficient of kinetic friction is 0.3, what is the magnitude of the frictional force exerted on an object weighing 4.5kg lying on an incline with angle \(\displaystyle 60^o\) to the horizontal?

 \(\displaystyle g=10 \frac{m}{s^2}\)

Possible Answers:

\(\displaystyle 2.5N\)

\(\displaystyle 5N\)

\(\displaystyle 7.5N\)

\(\displaystyle 15N\)

Correct answer:

\(\displaystyle 7.5N\)

Explanation:

Recall the formula to determine kinetic friction:

\(\displaystyle F_{kinetic}=\mu N\)

Here, \(\displaystyle F_{kinetic}\) is the kinetic friction force, \(\displaystyle \mu\) is the constant of friction, and \(\displaystyle N\) is the magnitude of the normal force of the object in question. Recall that the formula for the normal force is given by:

\(\displaystyle F_{normal}=mgcos(\theta)\)

Here, \(\displaystyle m\) is the mass of the object, \(\displaystyle g\) is the gravitational constant, and \(\displaystyle \theta\) is the angle of elevation. Since the object is lying \(\displaystyle 60^o\) to the horizontal, the normal force it feels is:

\(\displaystyle F_{normal}=4.5kg*10\frac{m}{s^2}*cos(60^o)= 4.5kg*10\frac{m}{s^2}*\frac{1}{2}\)

\(\displaystyle F_{normal}=22.5N\)

Therefore,

\(\displaystyle F_{kinetic}=0.3*22.5= 7.5N\)

Example Question #671 : Newtonian Mechanics

What are the units of the friction coefficient \(\displaystyle \mu\) in the formula:

\(\displaystyle F_{friction}=\mu\cdot F_{normal}\)

Possible Answers:

\(\displaystyle N\)

\(\displaystyle \frac{F\cdot kg}{m^2}\)

\(\displaystyle F^2\)

No units

\(\displaystyle F^{-1}\)

Correct answer:

No units

Explanation:

To solve for the units, we simply have to solve for \(\displaystyle \mu\) and determine its units. 

\(\displaystyle \mu=\frac{F_{friction}}{F_{normal}}\)

Both friction and normal forces have units of newtons \(\displaystyle N\), which means that \(\displaystyle \mu\) has units \(\displaystyle \frac{F}{F}=1\)

 

Therefore, \(\displaystyle \mu\) is a dimensionless term and has no units. 

Example Question #11 : Force Of Friction

A \(\displaystyle 15N\) crate slides along a horizontal platform at \(\displaystyle 5\frac{m}{s}\). The coefficient of kinetic friction between the crate and the platform is \(\displaystyle \mu=0.2\).

What is the magnitude of the force of friction exerted on the crate?

Possible Answers:

\(\displaystyle 3N\)

\(\displaystyle 0N\)

\(\displaystyle 5N\)

\(\displaystyle -5N\)

\(\displaystyle -3N\)

Correct answer:

\(\displaystyle 3N\)

Explanation:

The equation for frictional force is given by:

\(\displaystyle F_{f}=F_{N}*\mu\) 

Because the crate is on a horizontal platform, the normal force is equal to the weight force \(\displaystyle (F_N=15N)\). Multiply the normal force by the given coefficient of kinetic friction \(\displaystyle (\mu =0.2)\) and solve for the frictional force:

\(\displaystyle F_f=15N*0.2=3N\)

Note that the coefficient of friction is unitless because it is a ratio and therefore our answer comes out in Newtons. Also note that the force of friction is in a direction that opposes motion.

Example Question #111 : Specific Forces

A baseball player is running with a velocity of \(\displaystyle v=7 \frac{m}{s}\) and slides \(\displaystyle 5 m\) to make it to the home plate. Calculate the coefficient of kinetic friction, \(\displaystyle \mu _k\) between the player's clothes and the clay.

Possible Answers:

\(\displaystyle 0.075\)

\(\displaystyle 0.3\)

\(\displaystyle 0.15\)

\(\displaystyle 0.5\)

Correct answer:

\(\displaystyle 0.5\)

Explanation:

We can use one of the kinematic equations to our advantage: 

\(\displaystyle v_f^2=v_i^2+2a \Delta x\)

We are given the player's speed and the distance the player slide into home plate. What is implied is that the player's final velocity \(\displaystyle v_f=0\). We can also rewrite the acceleration \(\displaystyle a\) in terms of the player's mass and the forces acting on him (which is only friction)

\(\displaystyle a= \frac{F_f}{m}=\frac{-\mu _k m g}{m}= -\mu _k g\)

Substitute this into the kinematic equation and solve for the coefficient of kinetic friction, \(\displaystyle \mu _k\)

\(\displaystyle 0= (7 \frac{m}{s})^2+2(- \mu _k g) 5m \Rightarrow \mu _k = \frac{(7\frac{m}{s})^2}{2(5m)9.8\frac{m}{s^2}}=0.5\)

Example Question #11 : Force Of Friction

A car going \(\displaystyle 45 \frac{meters}{second}\) locks up it's brakes and stops in \(\displaystyle 30 m\). Determine the coefficient of friction of the tires on the road.

Possible Answers:

\(\displaystyle \mu=1.1\)

\(\displaystyle \mu=4.6\)

\(\displaystyle \mu=.33\)

\(\displaystyle \mu=7.3\)

\(\displaystyle \mu=3.4\)

Correct answer:

\(\displaystyle \mu=3.4\)

Explanation:

\(\displaystyle E_{initial}+\Delta E=E_{final}\)

Since the car is stopped at the end:

\(\displaystyle E_{initial}+\Delta E=0\)

\(\displaystyle KE_{initial}+F_{friction}*d=0\)

\(\displaystyle .5mv^2+mg\mu*d =0\)

Solve for coefficient of friction:

\(\displaystyle \mu=-\frac{.5v^2}{g*d}\)

Plug in values:

\(\displaystyle \mu=-\frac{.5(45)^2}{-9.8*30}\)

\(\displaystyle \mu=3.4\)

Example Question #11 : Force Of Friction

A heavy truck and a relatively light car with tires made of identical material are driving next to each other at \(\displaystyle 35\frac{m}{s}\). They both slam on the brakes at the exact same time. Which will stop first? Assume they each locked up the brakes and all friction was thus kinetic friction.

Possible Answers:

They will stop at the same time

The car

The truck

Cannot be determined without knowing the coefficient of kinetic friction

Correct answer:

They will stop at the same time

Explanation:

Use the following equations:

\(\displaystyle W=F*d\)

\(\displaystyle KE=.5mv^2\)

\(\displaystyle F_{friction}=\mu*F_{normal}=\mu*m*|g|\)

\(\displaystyle E_{intial}+W=E_{final}\)

The force of friction will point in the opposite direction as the direction of travel, so it will have a negative sign.

\(\displaystyle .5mv^2-\mu*m*|g|*d=0\)

Since the energy will be zero when the vehicle has stopped.

Solve for \(\displaystyle d\), the stopping distance.

\(\displaystyle d=\frac{.5v^2}{\mu*|g|}\)

\(\displaystyle m\), the mass is not in the solution. Indeed, the extra energy due to mass is cancelled out due to extra friction resulting from the higher mass.

Example Question #681 : Newtonian Mechanics

A small child struggles to pull his heavy suitcase through an airport. He tugs it along with a rope at an angle of \(\displaystyle 30^{\circ}\) to the ground. The child exerts a force of \(\displaystyle 50N\), which is just enough to pull the suitcase along with constant velocity. The coefficient of kinetic friction between the ground and the suitcase is \(\displaystyle \mu_k=0.3\)

What is the mass of the child's suitcase?

Possible Answers:

\(\displaystyle 17.28kg\)

\(\displaystyle 12.92kg\)

\(\displaystyle 8.50kg\)

\(\displaystyle 14.73kg\)

\(\displaystyle 12.18kg\)

Correct answer:

\(\displaystyle 17.28kg\)

Explanation:

Suitcase

The diagram above shows the free body diagram for the suitcase. For this problem, we take the positive y-direction to be the direction of the normal force and the negative x-direction to be the direction of the frictional force. Then applying Newton's 2nd law in the x and y-directions gives us the following equations:

\(\displaystyle \sum F_x=F_c\cos(30^\circ)-\mu_kF_N=ma_x\)

\(\displaystyle \sum F_y=F_N+F_c\cos(30^\circ)-mg=ma_y\)

Since the child pulls the suitcase at constant velocity, we have \(\displaystyle a_x=0\) by Newton's first law. Also, the object is constrained to the ground, which means that \(\displaystyle a_y=0\) as well. Thus we obtain:

\(\displaystyle F_c\cos(30^\circ)-\mu_kF_N=0\)

\(\displaystyle F_N+F_c\sin(30^\circ)-mg=0\)

We want to solve for \(\displaystyle \mu_k\), but we need to get rid of the normal force in the first equation. We can do so by solving for \(\displaystyle F_N\) in the second equation. Doing so, we obtain:

\(\displaystyle F_N=mg-F_c\sin(30^\circ)\)

Substituting this value into the first equation and solving for \(\displaystyle m\), we obtain:

\(\displaystyle F_c\cos(30^\circ)-\mu_k(mg-F_c\sin(30^\circ))=0\)

\(\displaystyle F_c\cos(30^\circ)-\mu_kmg+\mu_kF_c\sin(30^\circ)=0\)

\(\displaystyle F_c\cos(30^\circ)+\mu_kF_c\sin(30^\circ)=\mu_kmg\)

\(\displaystyle m=\frac{F_c\cos(30^\circ)+\mu_kF_c\sin(30^\circ)}{\mu_kg}\)

Substituting our unknowns into the expression on the right side, we get:

\(\displaystyle m=\frac{(50N)\cos(30^\circ)+(0.3)(50N)\sin(30^\circ)}{(0.3)(9.8\frac{m}{s^2})}=17.28kg\)

Therefore, the mass of the suitcase is \(\displaystyle 17.28kg\)

Example Question #11 : Force Of Friction

Vt physics 10 29 question 6

Suppose that we have zoomed in on the tire of a car touching the road, as shown in the figure. If the car moves to the right (towards A), what is the frictional force that acts on the tire due to the road?

Possible Answers:

Force of kinetic friction acting towards B

Force of static friction acting towards B

Force of kinetic friction acting towards A

Force of static friction acting towards A

Correct answer:

Force of static friction acting towards A

Explanation:

In this question, we're presented with a scenario in which we need to consider the frictional forces that occur between the tire of a moving vehicle and the road. More specifically, we'll need to determine the direction in which the frictional force acts, as well as whether the force is kinetic or static.

First, let's determine what type of force we're dealing with. Since the tire is rotating as the car itself moves forward, each point around the circumference of the tire will come into contact with the road. It's important to realize that since the surface of the tire is never moving relative to the road in which it is in contact with, this cannot be kinetic friction. Therefore, it is the static force of friction that the road provides in moving the car forward.

Next, let's see which direction the force is acting. If we imagine the car moving to the right towards A, we see that the car's tire is pushing against the road to the left. And because of Newton's third law, since the tire is pushing against the road towards the left, the road must be pushing back with an equal magnitude of force, but in the opposite direction, to the right. Thus, the frictional force of the road is both static and acting toward the right.

Example Question #721 : Ap Physics 1

Consider the following scenario:

Sledder

A sledder of mass \(\displaystyle m\) is at the stop of a sledding hill at height \(\displaystyle h\) with a slope of angle \(\displaystyle a\).

If the sledder is traveling down the slope with \(\displaystyle \measuredangle a = 35^{\circ}\) and accelerating at a rate of \(\displaystyle 5\frac{m}{s^2}\), what is the coefficient of kinetic friction between the sled and the snow? Neglect air resistance.

\(\displaystyle g = 10\frac{m}{s^2}\)

Possible Answers:

\(\displaystyle 0.09\)

\(\displaystyle 0.17\)

\(\displaystyle 0.21\)

\(\displaystyle 0.14\)

\(\displaystyle 0.06\)

Correct answer:

\(\displaystyle 0.09\)

Explanation:

We will begin with Newton's second law to solve this problem:

\(\displaystyle F_{net}=ma\)

There are two forces acting on the sledder in the direction of acceleration: gravity and friction. We will begin with gravity:

\(\displaystyle F_g = mg\)

However, this is a vertical force, and we need the component that is in the direction of acceleration. Therefore, we will multiply it by the sine function:

\(\displaystyle F_{g_{x}} = mgsin(a)\)

If you're wondering why sin was used, think about the situation practically. As the angle grows, the hill gets steeper, and there will be more gravitational force in the direction of acceleration. Hence, we use sine.

Moving on to the force of friction, we will use the FUN equation:

\(\displaystyle F_f = \mu_kN\)

Now we need an expression for the normal force, which is the gravitational force multiplied by cos:

\(\displaystyle N = mgcos(a)\)

Once again, if you're unsure why we used cosine, think about the situation practically. As the hill gets flatter (angle decreased), the normal force grows, and cos is the function that gets larger as the angle gets smaller. Now substituting this into the FUN equation, we get:

\(\displaystyle F_f = \mu_kmgcos(a)\)

Now combining our two forces (remember friction will be subtracted since it is in the opposite direction of movement):

\(\displaystyle mgsin(a)-\mu_kmgcos(a)=ma\)

Eliminating mass from the expression:

\(\displaystyle gsin(a)-\mu_kgcos(a)=a\)

And rearranging for the coefficient of friction:

\(\displaystyle \mu_k = \frac{gsin(a)-a}{gcos(a)}\)

We have each of these values, so we can solve:

\(\displaystyle \mu_k = \frac{\left ( 10\frac{m}{s^2}\right )sin(35^{\circ})-5\frac{m}{s^2}}{\left ( 10\frac{m}{s^2}\right )cos(35^{\circ})}\)

\(\displaystyle \mu_k = 0.09\)

Example Question #114 : Forces

Consider the following scenario:

Sledder

A sledder of mass \(\displaystyle m\) is at the stop of a sledding hill at height \(\displaystyle h\) with a slope of angle \(\displaystyle a\).

The sledder has a mass of \(\displaystyle 100kg\) and the hill has a slope of \(\displaystyle 30^{\circ}\). If the sledder has a velocity of \(\displaystyle 8\frac{m}{s}\) when he reaches the bottom of the hill, which becomes flat, and comes to a stop after \(\displaystyle 20m\), what is the coefficient of kinetic friction and the height of the hill? Neglect air resistance and any other frictional forces.

\(\displaystyle g = 10\frac{m}{s^2}\)

Possible Answers:

\(\displaystyle \mu_k = 0.16, h = 4.4m\)

\(\displaystyle \mu_k = 0.12, h = 15.1m\)

\(\displaystyle \mu_k = 0.08, h = 12.1m\)

\(\displaystyle \mu_k = 0.23, h = 7.2m\)

\(\displaystyle \mu_k = 0.02, h = 3.5m\)

Correct answer:

\(\displaystyle \mu_k = 0.16, h = 4.4m\)

Explanation:

This will be a two step problem. We need to calculate the coefficient of friction and height of hill. These will be calculated separately. So first, stop and think about which one we can calculate with the information we're given, and which one is dependent on the other.

From the problem statement, we can directly calculate the coefficient of kinetic friction using the velocity of the sledder at the bottom of the hill and the distance it takes for the sledder to come to rest. Since there is no height change, we know that all kinetic energy is dissipated as friction. Therefore, we can say:

\(\displaystyle KE = W_F\)

Expanding these terms:

\(\displaystyle \frac{1}{2}mv^2=\mu_kN\cdot d = \mu_k mgd\)

Eliminating mass:

\(\displaystyle \frac{1}{2}v^2 = \mu_k gd\)

Rearranging for the coefficient of friction:

\(\displaystyle \mu _k = \frac{v^2}{2gd}\)

Plugging in the values given in the problem statement:

\(\displaystyle \mu_k = \frac{\left ( 8\frac{m}{s}\right )^2}{2\left ( 10\frac{m}{s^2}\right )(20m)}\)

\(\displaystyle \mu_k =0.16\)

Now that we have our coefficient of friction, we can move on calculating the height of the hill. For this we can use the expression for conservation of energy:

\(\displaystyle E = U_i+K_i = U_f+K_f +W\)

The sledder is at rest at the top of the hill, so we can eliminate initial kinetic energy. Also, if we assume that the bottom of the hill has a height of 0, we can eliminate final potential energy to get:

\(\displaystyle U_i = K_f + W\)

Expanding each term one at a time from left to right:

\(\displaystyle U_i = mgh\)

\(\displaystyle K_f = \frac{1}{2}mv_f^2\)

\(\displaystyle W = F_fd= \mu_k Nd\)

We need to substitute expression for the normal force and distance that the frictional force is applied. For the normal force we get:

\(\displaystyle N = mgcos(\theta)\)

And for distance, we begin with the sine function:

\(\displaystyle sin(\theta)= \frac{h}{d}\)

Rearranging for d, we get:

\(\displaystyle d = \frac{h}{sin(\theta)}\)

Substituting these into the expression for work, we get:

\(\displaystyle W = \mu_k mgcos(\theta)\left ( \frac{h}{sin\theta}\right )\)

Cleaning up the equation a bit:

\(\displaystyle W = \mu_kmghcot(\theta)\)   (4)

Now substituting the expanded expressions back into the first expression, we get:

\(\displaystyle mgh = \frac{1}{2}mv_f^2+\mu_kmghcot(\theta)\)

We can eliminate mass to get:

\(\displaystyle gh = \frac{1}{2}v_f^2+\mu_kghcot(\theta)\)

Now we need to rearrange for height:

\(\displaystyle gh -\mu_kghcot(\theta)= \frac{1}{2}v_f^2\)

\(\displaystyle h\left (g -\mu_kgcot(\theta) \right )= \frac{1}{2}v_f^2\)

\(\displaystyle h= \frac{v_f^2}{2 \left (g -\mu_kgcot(\theta) \right )}\)

\(\displaystyle h= \frac{v_f^2}{2g \left ( 1-\mu_kcot(\theta) \right )}\)

We have all of these values, so time to plug and chug:

\(\displaystyle h = \frac{\left ( 8\frac{m}{s}\right )^2}{2 \left (10\frac{m}{s^2} \right )\left ( 1-0.16cot(30^{\circ})\right )}\)

\(\displaystyle h = 4.4m\)

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