Newton's Laws Applications Review
Navigate to Answers for:
Part D: Problem-Solving
22. A 945-kg car traveling rightward at 22.6 m/s slams on the brakes and skids to a stop (with locked wheels). If the coefficient of friction between tires and road is 0.972, determine the distance required to stop. PSYW
Answer: 26.8 m
Like most problems, this problem begins with a free-body diagram (as shown at right). Note that there is no rightwards applied force (a common mistake). Note also that the force of friction is the only force responsible for the acceleration (deceleration) of the car. The Ffrict value is the net force. So determining the acceleration involves finding Fgrav (Fgrav=m•g = 945•9.8 = 9261 N), Fnorm (the same as Fgrav= 9261 N), and Ffrict (Ffrict = mu•Fnorm = 0.972 • 9261 N = 9002 N). With the ∑Fx = 9002 N, the acceleration can be calculated (ax = ∑Fx/m = 9002 N/945 kg = 9.53 m/s/s.) This is a leftwards acceleration; ax will be assigned the numerical value of -9.53 m/s/s in the next part of this problem.
Now in the kinematic part of this problem, the distance must be found using the known information (vf = 0 m/s, vo = 22.6 m/s and a = 9.53 m/s/s). Use the equation:
(0 m/s)2 = (22.6 m/s)2 + 2•(-9.53 m/s/s)•d
d = [(22.6 m/s)2] / [2•(9.53 m/s/s)] = 26.8 m
Useful Web Links
Kinematic Equations and Problem-Solving || Finding Individual Forces
23. A student pulls a 2-kg backpack across the ice (assume frictionless) by pulling at a 30.0 degrees angle to the horizontal. The velocity-time graph for the motion is shown. Perform a careful analysis of the situation and determine the applied force. PSYW
Answer: 0.289 N
The acceleration of the object can be found from the slope of the line on the v-t graph. The acceleration is:
The net force is found from m•a; so Fnet = (2 kg)•(0.125 m/s/s) = 0.250 N.
The only force that can contribute to the net force is the applied force. The horizontal component of the applied force must therefore be equal to 0.250 N. A diagram depicting the relationship of the lengths of the side for this 30-60-90 triangle is shown at the right. The cosine function can be utilized to determine the magnitude of the tension force.
Solving for Fapp yields 0.289 N.
Net Force Problems Revisited || The Meaning of Shape for a v-t Graph
Answer: 56.6 N
If the mass of the sign is 10.0 kg, then the weight of the sign is 98 N (Fgrav = m•g). This downward force of gravity must be balanced by the upward component of the tension. Thus, each cable must pull upwards with 49.0 N of force. In addition to the upward pull by the cable, there is also a horizontal pull. The tension force is the combined effect of these two componens. To determine the tension, set up a 60-30-90 triangle with a vertical component of 4.09 N as shown at the right. Then, use the sine function to solve for the hypotenuse of the triangle. The work is shown below:
Ftens = (49.0 N)/(sin 60 deg) = 56.6 N
Equilibrium and Statics
25. Splash Mountain at Disney World in Orlando, Florida is one of the steepest water plume rides in the United States. Occupants of the boat fall from a height of 100 feet (3.2 ft = 1 m) down a wet ramp which makes a 45 degree angle with the horizontal. Consider the mass of the boat and its occupants to be 1500 kg. The coefficient of friction between the boat and the ramp is 0.10. Determine the frictional force, the acceleration, the distance traveled along the incline, and the final velocity of the boat at the bottom of the incline. PSYW
Answer: All numerical answers are rounded to two significant digits: Ffrict = ~1040 N; a = ~6.2 m/s/s; d = ~44 m; vf = 23 m/s
This is a multi-part inclined plane problem. Like all problems, it should begin with a free-body diagram. If this is where the difficulty lies for you, then take some time to review inclined planes. See help page. The friction force opposes the motion of the boat. If the + x direction is defined as down and along the incline (as shown at the right) then the net force can be computed by the expression F|| - Ffrict. The parallel component of the weight vector (m•g•sin theta) is 10395 N. The Ffrict force is computed by multiplying the coefficient of friction (0.10) by the normal force (Fnorm). The normal force is balancing Fperp (m•g•cos theta) and is equal to it. So Fnorm = 10395 N and Ffrict = 1039.4 N. The net force (Fnet = F|| - Ffrict ) is 9356 N; and the acceleration is Fnet/m or 6.24 m/s/s.
Once the Newton's laws analysis has been completed and the acceleration determined, the kinematics portion of the problem can be tackled. First, determine the distance along the incline from the height of the hill and the angle of incline. The height of 100 ft is equivalent to 31.25 m; the distance along the incline is 44.19 m [found from (31.25 m)/sin (45)]. The final velocity is found using the acceleration, distance and a kinematic equation:
vf2 = (0 m/s)2 + 2•(6.24 m/s/s)•(31.25 m) = 551.28 m2/s2
vf = 23 m/s
Kinematic Equations and Problem-Solving || Inclined Planes
26. At last year's Homecoming Pep Rally, Trudy U. Skool (attempting to generate a little excitement) slid down a 42.0 degree incline from the sports dome to the courtyard below. The coefficient of friction between Trudy's jeans and the incline was 0.650. Determine Trudy's acceleration along the incline. Begin with a free-body diagram. PSYW
Answer: 1.82 m/s/s
This is nearly an identical force analysis as the last problem except the mass is not known. When the mass is not known, the problem-solving strategy involves inserting m into the equation as an unknown variable and proceding with the solution. It is likely that in a subsequent step of the problem that the m will cancel and the accceleration can be determined without knowing m.
The net force is found by the expression F|| - Ffrict. The parallel component of the weight vector is m•g•sin(theta). The Ffrict force is computed by multiplying the coefficient of friction (mu) by the normal force (Fnorm). The normal force is balancing Fperp and is equal to m•g•cos(theta). So Fnorm = m•g•cos(theta) and Ffrict = µ•m•g•cos(theta). The net force is m•g•sin(theta) - µ•m•g•cos(theta). The acceleration is Fnet/m or [m•g•sin(theta) - µ•m•g•cos(theta)]/m. Note that there is an m in both terms in the numerator and a m in the denominator. The masses cancel and the equation reduces to
By substitution of the given values into the equation, it can be shown that a = 1.82 m/s/s.
27. Baldwin Young is
conducting his famous toupee
experiments. He tips his head at a given angle and determines the
coefficient of static and kinetic friction between a toupee (which is
probably his own) and his scalp. At an angle of just barely 17.5
degrees, the toupee begins to accelerate from rest. Then Baldwin
lowers the angle to 13.8 degrees to observe that the toupee moves
with a constant speed. Use this information to determine the
coefficients of both static and kinetic friction. Begin with a
free-body diagram. PSYW
Answer: µstatic = 0.315; µkinetic = 0.246
A free-body diagram and a force analysis will yield the equation
for both the static and the kinetic situation. That is, when the object is at rest or moving down the incline at constant speed, the force down and parallel to the incline is balanced by the force which is up and parallel to the incline. The expressions for Ffrict and F|| can be substituted into this equation to yield
Knowing that Fnorm is equal in magnitude to Fperp, the expression for Fperp can be substituted into the equation for Fnorm. The new equation becomes
The angles for both the static and the kinetic case are known and mu (coefficient of friction) is the unknown. So the equation can be re-arranged to
The mass (m) and the acceleration of gravity (g) cancel from both the numerator and the denominator leaving sin(theta)/cos(theta) on the right side. This can be simplified again to
Equation 1 can be used to find the coefficient of friction for both the static and kinetic situation. The answers are as follows:
µkinetic = tan(13.8 deg.) = 0.246
28. Consider the two-body system at the right. There is a 0.250-kg object accelerating across a rough surface. The sliding object is attached by a string to a 0.100 kg object which is suspended over a pulley. The coefficient of kinetic friction is 0.183. Calculate the acceleration of the block and the tension in the string. PSYW
Answer: a = 1.52 m/s/s ; Ftens = 0.83 N
The solution to this problem begins by drawing a free-body diagram for each object. Note that the + x-axis for the 0.25-kg object is drawn in the direction that the object accelerates; and the + y-axis for the 0.10-kg object is drawn in the direction which it accelerates.
Note that while friction acts upon the 0.25-kg object, it does not act upon the 0.10-kg object (since it is not being dragged across the surface). Newton's second law of motion (·F = m•a) can be applied to the motion of the 0.25-kg mass:
Ftens - Ffrict = m•ax
The expressioin for Ffrict can be substituted into the equation.
Since the vertical forces on the 0.250-kg object balance each other, it is known that Fgrav = Fnorm; so Fnorm = m•g = (0.250 kg)•(9.8 m/s/s) = 2.45 N. Since µ = 0.183, µ•Fnorm = 0.448 N. Equation 1 can now be re-written as
Newton's second law of motion (∑F = m•a) can be applied to the motion of the 0.10o-kg mass:
Fgrav - Ftens = m•ay
The expression for Fgrav (m•g) can be substituted into the equation.
(0.100 kg)•(9.8 m/s/s) - Ftens = m•ay
0.98 N - Ftens = m•ay
The mass of the 0.100-kg object can be substituted into the equation to yield equation 3.
Equations 2 and 3 each include an acceleration term - ax and ay. These acceleration values are the same since the system accelerates together; thus, ax = ay = a. An inspection of these two equations show that there are now two equations and two unknowns - Ftens and a. These unknown values can be solved for in the customary manner.
First, equation 3 is rearranged to create an expression for Ftens:
This expression for Ftens is now substituted into equation 2.
0.980 N - (0.100 kg)•a - 0.448 N = (0.250 kg)•a
This equation can now be solved for acceleration (a) as shown in the following steps.
0.532 N = (0.305 kg)•a
a = 1.52 m/s/s
Now that a is known, its value can be substituted into equation 4 in order to solve for Ftens.
0.980 N - (0.10 kg)•(1.52 m/s/s) = Ftens
0.980 N - 0. 152 N = Ftens
0.830 N = Ftens
29. Consider the two-body system at the right. A 22.7-N block is placed upon an inclined plane which is inclined at a 17.2 degree angle. The block is attached by a string to a 34.5-N block which is suspended over a pulley. The coefficient of friction is 0.219. Determine the acceleration of the block and the tension in the string. PSYW
Answer: a = 3.95 m/s/s; Ftens = 20.6 N
Like #21 and #28, this problem can most easily be solved using separate free-body analyses on the individual masses. Free-body diagrams, the chosen axes systems, and associated information is shown below. Note that in chosing the axis system, it has been assumed that object 1 will accelerate up the hill and object 2 will accelerate downwards on the back side of the hill. If this ends up to be false, then the acceleration values will turn out to be negative values.
The 22.7-N object is on an inclined plane. The usual circumstances apply; there is no acceleration along what has been designated as the y-axis.
The parallel component of Fgrav is
F|| = 6.71 N
The Ffrict value can be found from the Fnorm value:
The ∑Fx = m•ax equation can now be written:
Ftens - Ffrict - F|| = m•ax
Ftens - 4.75 N - 6.71 N = (2.32 kg)•ax
(Note that the Ffrict and F|| forces are subtracted from Ftens since they are heading in the direction of the negative x-axis.)
The above process can be repeated for the 34.5-N object. The ∑Fy = m•ay equation can now be written:
Fgrav - Ftens = m•ay
(34.5 N) - Ftens = (3.52 kg)•ay
The separate free-body analyses have provided two equations with two unknowns; the task at hand is to use these two equations to solve for Ftens and a.
Equation 2 can be re-written as
Since both objects accelerate together at the same rate, the ax for the 22.7-N object is equal to the ay value for the 34.5-N object. The subscripts x and y can be dropped and a can be inserted into each equation.
Equation 3 provides an expression for Ftens in terms of a. This expression is inserted into equation 1 in order to solve for acceleration. The steps are shown below.
(34.5 N) - (3.52 kg) •a - 4.75 N - 6.71 N = (2.32 kg)•a
23.0 N = (5.84 kg)•a
3.95 m/s2 = a
The value of a can be re-inserted into equation 3 in order to solve for Ftens:
Ftens = 20.6 N
30. A 10.0-gram mass is tied to a string. The string is attached to a 500.0-gram mass and stretched over a pulley, leaving the 10.0-gram mass suspended above the floor. Determine the time it will take the 10.0-gram mass to fall a distance of 1.50 meters if starting from rest. (No friction)
Answer: t = 3.95 s
This problem involves a blending of Newton's laws and kinematics. Newton's laws will have to be used to determine the acceleration; then kinematics will have to be used to determine the time to fall 1.50 m from rest. The solution to this problem begins by drawing a free-body diagram for each object. Note that the + x-axis for the 500.0-gram object is drawn in the direction that the object accelerates; and the + y-axis for the 10.0-gram object is drawn in the direction which it accelerates.
Note that while friction acts upon the 500.0-gram object, it does not act upon the 10.0-gram object (since it is not being dragged across the surface). Newton's second law of motion (∑F = m•a) can be applied to the motion of the 0.5000-kg mass:
Ftens = m•ax
Ftens = (0.5000 kg)•ax
Newton's second law of motion (∑F = m•a) can also be applied to the motion of the 0.0100-kg hanging mass:
Fgrav - Ftens = m•ay
The expression for Fgrav (m•g) can be substituted into the equation.
(0.0100 kg)•(9.8 m/s/s) - Ftens = m•ay
0.098 N - Ftens = m•ay
The mass of the 0.0100-kg object can be substituted into the equation to yield equation 2.
Equations 1 and 2 include an acceleration term - ax and ay. These acceleration values are the same since the system accelerates together; thus, ax = ay = a. An inspection of these two equations show that there are now two equations and two unknowns - Ftens and a. These unknown values can be solved for in the customary manner. First, equation 2 is rearranged so as to express Ftens in terms of a. This yields equation 3.
Ftens = 0.098 N - (0.0100 kg)•ayEquations 1 and 3 both express Ftens in terms of a. The right side of each equation is equal to Ftens ; so these two right sides can be set equal to each other to yield the following equation.
(0.5000 kg)•ax = 0.098 N - (0.0100 kg)•ay
Since ax is equal to ay, the equation can be re-written as
(0.5000 kg)•a = 0.098 N - (0.0100 kg)•a
where a is the horizontal and the vertical acceleration of the 500.0-gram and 10.0-gram masses. Now the above equation can be manipulated in order to solve for the value of a.
(0.5000 kg)•a + (0.0100 kg)•a = 0.098 N
(0.510 kg)•a = 0.098 N
a = (0.098 N) / (0.510 kg)
a = 0.1922 m/s/s
Now that an acceleration has been determined, the time can be calculated using a kinematic equation. The three known kinematic quantities are:
vi = 0 m/s
a = 0.1922 m/s/s
d = 1.50 m
The following kinematic equation can be used to solve for time (t). The substitutions and algebra are shown.
d = vi•t + 0.5 •a• t2
1.50 m = (0 m/s) • t + 0.5•(.1922 m/s/s)•t2
1.50 m = (0.0961 m/s/s)•t2
t2 = (1.50 m) / (0.0961 m/s/s)
t2 = 15.6087 s2
t = SQRT(15.6087 s2)
t = 3.95 s
Kinematic Equations and Problem-Solving