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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 F_frict value is the net force. So determining the acceleration involves finding F_grav (F_grav=m•g = 945•9.8 = 9261 N), F_norm (the same as F_grav= 9261 N), and F_frict (F_frict = mu•F_norm = 0.972 • 9261 N = 9002 N). With the ∑F_x = 9002 N, the acceleration can be calculated (a_x = ∑F_x/m = 9002 N/945 kg = 9.53 m/s/s.) This is a leftwards acceleration; a_x 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 (v_f = 0 m/s, v_o = 22.6 m/s and a = 9.53 m/s/s). Use the equation:

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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 F_net = (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 F_app yields 0.289 N.

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Answer: 56.6 N

If the mass of the sign is 10.0 kg, then the weight of the sign is 98 N (F_grav = 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:

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Answer: All numerical answers are rounded to two significant digits:  F_frict = ~1040 N; a = ~6.2 m/s/s; d = ~44 m; v_f = 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_|| - F_frict. The parallel component of the weight vector (m•g•sin theta) is 10395 N. The F_frict force is computed by multiplying the coefficient of friction (0.10) by the normal force (F_norm). The normal force is balancing F_perp (m•g•cos theta) and is equal to it. So F_norm = 10395 N and F_frict = 1039.4 N. The net force (F_net = F_|| - F_frict ) is 9356 N; and the acceleration is F_net/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:

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Answer: µ_static = 0.315; µ_kinetic = 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 F_frict and F_|| can be substituted into this equation to yield

Knowing that F_norm is equal in magnitude to F_perp, the expression for F_perp can be substituted into the equation for F_norm. 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

µ = tan(theta)

Equation 1 can be used to find the coefficient of friction for both the static and kinetic situation. The answers are as follows:

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Answer: a = 1.52 m/s/s ; F_tens = 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:

The expressioin for F_frict can be substituted into the equation.

F_tens - µ•F_norm = m•a_x

Since the vertical forces on the 0.250-kg object balance each other, it is known that F_grav = F_norm; so F_norm = m•g = (0.250 kg)•(9.8 m/s/s) = 2.45 N. Since µ = 0.183, µ•F_norm = 0.448 N. Equation 1 can now be re-written as

F_tens - 0.448 N = (0.250 kg)•a_x

Newton's second law of motion (∑F = m•a) can be applied to the motion of the 0.10o-kg mass:

The expression for F_grav (m•g) can be substituted into the equation.

The mass of the 0.100-kg object can be substituted into the equation to yield equation 3.

0.98 N - F_tens = (0.100 kg)•a_y

Equations 2 and 3 each include an acceleration term - a_x and a_y. These acceleration values are the same since the system accelerates together; thus, a_x = a_y = a. An inspection of these two equations show that there are now two equations and two unknowns - F_tens and a. These unknown values can be solved for in the customary manner.

First, equation 3 is rearranged to create an expression for F_tens:

0.98 N - (0.100 kg)•a = F_tens

This expression for F_tens is now substituted into equation 2.

F_tens - 0.448 N = (0.250 kg)•a

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.

Now that a is known, its value can be substituted into equation 4 in order to solve for F_tens.

Answer: a = 3.95 m/s/s; F_tens = 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 F_grav is

The F_frict value can be found from the F_norm value:

The ∑F_x = m•a_x equation can now be written:

(Note that the F_frict and F_|| forces are subtracted from F_tens 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 ∑F_y = m•a_y equation can now be written:

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 F_tens and a.

Equation 2 can be re-written as

Since both objects accelerate together at the same rate, the a_x for the 22.7-N object is equal to the a_y value for the 34.5-N object. The subscripts x and y can be dropped and a can be inserted into each equation.

(34.5 N) - (3.52 kg) •a = F_tens

Equation 3 provides an expression for F_tens in terms of a. This expression is inserted into equation 1 in order to solve for acceleration. The steps are shown below.

The value of a can be re-inserted into equation 3 in order to solve for F_tens:

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:

Newton's second law of motion (∑F = m•a) can also be applied to the motion of the 0.0100-kg hanging mass:

The expression for F_grav (m•g) can be substituted into the equation.

The mass of the 0.0100-kg object can be substituted into the equation to yield equation 2.

0.098 N - F_tens = (0.0100 kg)•a_y

Equations 1 and 2 include an acceleration term - a_x and a_y. These acceleration values are the same since the system accelerates together; thus, a_x = a_y = a. An inspection of these two equations show that there are now two equations and two unknowns - F_tens and a. These unknown values can be solved for in the customary manner. First, equation 2 is rearranged so as to express F_tens in terms of a.  This yields equation 3.

F_tens = 0.098 N - (0.0100 kg)•a_y

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