You are driving your car along a level residential street at 35 mph (16 m/s) when you see a ball roll out into the street in front of you. You slam on your brakes in case anyone follows the ball out into the street. If the coefficient of friction between your car and the road is 0.67, how far do you travel before you come to a stop?
In this problem, you are asked to relate motion (the car comes to a stop) to force (friction). Force and motion of a single object are always related through Newton’s Second Law, so this is a force or 2nd Law problem even though you are asked about distance! You should always approach a problem first by thinking about the key interactions described, regardless of what quantity you are asked to find. Going from acceleration to distance will be a second step in this problem.
If you have already studied energy, you will recognize that this problem can be solved more efficiently using energy. Click here for the energy solution.
The final step of this problem will, indeed, involve kinematics. If you recognize the kinematic question before you recognize the 2nd Law nature of the problem, you will see that you need acceleration in order to solve you equation. That will prompt you to approach the problem from Newton’s Laws and then bring the value for acceleration back into the kinematic expression. So you will work the same steps but in slightly different order.
Yes, if you are already at that point in your book. You will still need to do a 2nd Law problem to solve for force, but then can put it into the Conservation of Energy equation.
The free body diagram is a stylized drawing to help you visualize the cause of acceleration, and to directly map the drawing into the left hand side (ΣF) of Newton’s Second Law. The changing velocity (acceleration) is the effect of the forces, and therefore is not included on the list of forces.
Although the car initially has a velocity (to the right as I drew the picture), there is not a net force in the direction of motion. If you drive a car at a constant speed, the forward force between the tires and the road balances the resistive forces acting to slow the car down. In this case, you are braking the car and so the force between the car and the road is opposite to the direction of motion. The car continues forward because of inertia. Velocity is not a force and is not shown on the FBD.
All objects near Earth experience a downward force of gravity equal to mass of the object x 9.8 m/s^{2}. In this case, you are not told the mass of the car. However, the other forces present also depend on mass and so it will divide out as you solve the equations. You are not expected to recognize that as you begin the problem. As long as you approach a problem based on the key interactions (in this case, forces), you don’t need to think about numbers or even what quantity you are solving for until it is time to actually fill into the equations.
You do not need to draw a picture of the stopped car further down the road as you begin this problem. For many students, however, it helps to include that picture because you know it will become important later. So it is fine, but not required, to include it now. Regardless, note that it does not in any way affect how the FBD is drawn or how Newton’s 2nd Law will be set up.
The force of gravity acts downward, which is the negative y-direction. Therefore, mg is given a negative sign in the equation. Do not also give 9.8 m/s^{2} a negative sign, or you will double count the direction.
Even though this is a 1-dimensional kinematics problem, a, v, and x are vectors and so signs must be included in the components. Velocity is to the right in this case and so v’s were given + signs. Acceleration is to the left, and so a is negative. The positive sign on Δx tells you that the final location is to the right of the initial location.
The subscript “0” on v_{0} refers to the point in the motion that is earlier in time, or in this case when the car moves at 16 m/s. Therefore, v_{0} = 16 m/s, and v = 0.
Knowing that the street is level—that the car isn’t traveling on a hill tells you that normal force is straight up and that friction acts horizontally.
Although the car initially has a velocity (to the right as I drew the picture), there is not a net force in the direction of motion. If you drive a car at a constant speed, the forward force between the tires and the road balances the resistive forces acting to slow the car down. In this case, you are braking the car and so the force between the car and the road is opposite to the direction of motion. The car continues forward because of inertia. Velocity is not a force and is not shown on the FBD.
Braking force acts against the direction of motion, so to the left as I have drawn the picture. Because the road is horizontal, and braking causes an acceleration (change in velocity) along the road, you should also pick a coordinate system in which one axis is horizontal along the road. The math is always easier when you choose one axis to be in the direction of acceleration.
You do not need to draw a picture of the stopped car further down the road as you begin this problem. For many students, however, it helps to include that picture because you know it will become important later. So it is fine, but not required, to include it now. Regardless, note that it does not in any way affect how the FBD is drawn or how Newton’s 2nd Law will be set up.
All objects near Earth experience a downward force of gravity equal to mass of the object x 9.8 m/s^{2}. In this case, you are not told the mass of the car. However, the other forces present also depend on mass and so it will divide out as you solve the equations. You are not expected to recognize that as you begin the problem. As long as you approach a problem based on the key interactions (in this case, forces), you don’t need to think about numbers or even what quantity you are solving for until it is time to actually fill into the equations.
The key equation for any problem that relates forces and motion is Newton’s Second Law. Regardless of what quantity you are asked to find, begin with the Second Law. If additional information is needed, it will become apparent as you proceed even when that additional information is the quantity you are asked to find.
Step 1
In this problem, you are asked to find the stopping distance of the car. Because you were given information about the strength of the braking force, you were able to find the acceleration of the car. Acceleration can now be used to look at variables as distance, which describe the motion. Scroll down to continue to the second part of this problem.
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Step 2
At this stage in the problem, you know acceleration and velocities and are asked to find distance. In other words, you are asked to describe motion, or do a kinematics problem.
To select the appropriate kinematic equation, recognize that you are asked for the location when the car has stopped. In other words, you are asked to relate position (x) and velocity (v).
v^{2} = v_{0}^{2} + 2a(Δx)
0 = (16 m/s)^{2} + 2(-6.6 m/s^{2}) (Δx)
-256 m^{2}/s^{2} = -13.2 m/s^{2}(Δx)
19 m = Δx
y-component forces are normal force (up, or positive direction) and gravity (down, or negative direction.)
The car does not move in or out of the road, but only along it. There is no acceleration (changing velocity) in the y-direction.
Friction is to the left as I drew the diagram, acting in the opposite direction as the motion of the car. Left was chosen to be the –x direction, and so friction must be given a – sign in the equation.
The acceleration of the car is in the x-direction. I dropped the subscript, because a_{x} = a. (Note this is a statement about magnitude. The direction of a, to the left, will come out as you solve the equation.)
In this case, normal force is equal to mg. That is not always true (for example, if you are on incline, or accelerating vertically, or have other y-direction forces.) Do not assume that n = mg, but rather solve for n in the Fy equation .
Slamming on your brakes implies that your brakes are locked or at least that you are using the maximum frictional force available to you. Therefore, you can take friction to be at its maximum value of µn.
The subscript “0” on v_{0} refers to the point in the motion that is earlier in time, or in this case when the car moves at 16 m/s. Therefore, v_{0} = 16 m/s, and v = 0.
Even though this is a 1-dimensional kinematics problem, a, v, and x are vectors and so signs must be included in the components. Velocity is to the right in this case and so v’s were given + signs. Acceleration is to the left, and so a is negative. The positive sign on Δx tells you that the final location is to the right of the initial location.
The final step of this problem will, indeed, involve kinematics. If you recognize the kinematic question before you recognize the 2nd Law nature of the problem, you will see that you need acceleration in order to solve you equation. That will prompt you to approach the problem from Newton’s Laws and then bring the value for acceleration back into the kinematic expression. So you will work the same steps but in slightly different order.
Yes, if you are already at that point in your book. You will still need to do a 2nd Law problem to solve for force, but then can put it into the Conservation of Energy equation.
The free body diagram is a stylized drawing to help you visualize the cause of acceleration, and to directly map the drawing into the left hand side (ΣF) of Newton’s Second Law. The changing velocity (acceleration) is the effect of the forces, and therefore is not included on the list of forces.
All three forces present depend on mass, as you can see once you set up the equations. Therefore, it will divide out in the solution. In general, if you only have n, fr, and mg as forces, the answer will be independent of mass.
There is! However, air resistance is likely to be much less than the braking force, especially as the car slows down. The problem does not provide any information that you need to include air resistance, and so provides the hint that it is small. Think about this: the stopping distance of your car if you just take your foot off of the accelerator is much larger than the stopping distance when you slam on the brakes.
It is true that air puts a small buoyant force on the car. However, the size of the upward buoyant force compared to the downward force of gravity is very small. (The ratio of the two is given by the ratio of the density of air to the density of the car.) So to the number of significant figures included in this problem, we do not need to take buoyant force into account.
The force of gravity acts downward, which is the negative y-direction. Therefore, mg is given a negative sign in the equation. Do not also give 9.8 m/s
ΣF=ma
ΣF_{x}=ma_{x}
-fr = ma
-6.6 m/s^{2} = a
In this problem, the car comes to a stop (accelerates) because of the frictional force. Therefore, Newton’s 2^{nd} Law was used to find the acceleration on the car. The negative sign on 'a' tells you that acceleration is opposite to the (+ direction) motion of the car—friction is acting to slow the car down. The relatively high value (about 2/3 the acceleration due to gravity) indicates that, indeed, you slammed on the brakes.
Once we know the acceleration that results from the forces present, we are able to describe the motion. Because we are asked to relate velocity and location, the most productive equation is v^{2} = v_{0}^{2} + 2a(Δx), which gives a stopping distance of 19 m. City blocks are typically 100-200 m, so this is a stopping distance of about 1/5 of a block or less.