 Phyllis Fleming Physics Physics 104
 Answers - Newton's Laws

1.

The slope of the velocity versus time graph is the acceleration (Fig 1 below). The acceleration for:

(a) object A = (4 - 0)m/s/(4.0 - 0)s = 1.0 m/s2

(b) object B = (2 - 0)m/s/(4.0 - 0)s = 0.5 m/s2.

(c) mB/mA = aA/aB = 1.0/0.5 = 2. mB = 2mA = 2(2.0 kg) = 4.0 kg.

(d) Fnet = mAaA = mBaB = (2.0 kg)(1.0 m/s2) = (4.0 kg)(0.5 m/s2) = 2.0 N

2.

We again use the Principle of Superposition in solving this problem by separating the problem into forces acting along the horizontal or X-axis and those acting along the vertical or Y-axis. Then all forces will be along a line. We assign a positive or negative sign to the forces, depending on their direction along the axis, and then add them algebraically. There is no acceleration in the Y-direction, ay = 0.

 (Fnet)y = may = m(0) = 0 FN - Fg = 0 or FN = Fg = mg

(a) and (b). For m = 2.0 kg, FN = Fg = (2.0 kg)(10 m/s2) = 20 N
(Fnet)x = max

(a) For F = 12 N, F = max or ax = F/m = 12 N/2.0 kg= 6.0 m/s2

(b) For F =24 N,   ax = 24 N/2.0 kg = 12.0 m/s2
The acceleration is directly proportional to the net force. When you double the net force, you double the acceleration.

(c) For m = 1.0 kg, FN = Fg = (1.0 kg)(10 m/s2) = 10 N.
ax = 12 N/1.0 kg = 12 m/s2
The acceleration is inversely proportional to the mass of the object. When you halve the mass, you double the acceleration.

3.

Now the net force in the X-direction depends on an applied force to the right and a frictional force to the left.

 (Fnet)x = max F - f = max (a) 12 N - 4 N = 2.0 kg ax 4.0 m/s2 = ax (b) 24 N - 4 N = 2.0 kg ax 10 m/s2 = ax

While the applied force F is doubled, the net force is not doubled and the acceleration is not doubled. The second net force = 20/8 x the first net force = 5/2 the first net force. The second acceleration = 5/2 the first acceleration
= 5/2 x 4.0 m/s2 = 10 m/s2.

4.

In Fig. 4a below, I have isolated (shown by dashed box) the entire system in order to find the acceleration of the two blocks. In Fig. 4b below, I have isolated A and B separately in order to find FAB and FBA, the force on A due to B and the force on B due to A, respectively. FAB is an external force to the system which consists only of A and FBA is an external force to the system which consists only of B. FAB and FBA are internal forces to the system of Fig. 4a. I do not use a Y-axis in the figures because I am not interested in the normal force and there is no friction in the problem. For the entire system of Fig. 4a, I know the external force that acts on it along the X-axis and I know the mass of the system so I can find its acceleration. Since I am interested only in the X-direction, I drop the subscript x on Fnet and a.

 (a) Fnet external = (mass of system)a 12 N = (2 + 4)kg (a), or 12 N/6 kg = 2 m/s2 = a

Now we isolate block A to find FAB since we know mA and a and then isolate block B because we know everything but FBA, as shown in Fig. 4b. F is only in contact with A and A pushes B to the right, while B pushes A to the left.

 For block A, For block B, (b) (Fnet external)on A = (mA)a (c) (Fnet external)on B = (mB)a F - FAB = (2 kg)(2 m/s2) FBA = (4 kg)(2 m/s2) 12 N - FAB = 4 N or FAB = 8 N FBA = 8 N

FAB and FBA are Newton third law of motion forces. They are equal in magnitude, but opposite in direction. FAB, the force of B on A, has a magnitude of 8 N and is to the left. FBA, the force of A on B, has a magnitude of 8 N and is to the right.

5.

In the figure below, FPF represents the force of the floor on the person, FPR the force of the rope on the person, FRP the force of the person on the rope, FRH the force of the hook on the rope, FHR the force of the rope on the hook, and FHW the force of the wall on the hook. The Newton third law of motion forces are: (1) FRP and FPR, (2) FRH and FHR. 6.

It is a little difficult to solve (a) of this problem without a sneak look at the figure for part (b).  So, both figures are shown immediately below: Fig. for #6a shows only the external force in the horizontal direction acting on the system of the two blocks. Again Fnet external = ma, where a is the acceleration of the blocks and m is the mass of the system we have isolated. There are internal forces acting in the system, for example, the force of the rope on A, FAR, the force of the rope on B, FBR. And we cannot forget the force of A on the rope, FRA, or the force of B on the rope, FRB. I have not drawn the vectors for the gravitational forces on the blocks or the normal force of the surface on the blocks because they were not asked for and there is no frictional force. While Figures for #6b and c show forces FAR, FBR, FRA and FRB, these are internal to the system shown in Fig. for #6a. The only external force for the entire system in the horizontal direction is F = 12 N.

(a) Thus for the system shown in Fig. for #6a:

 (Fnet external)x = (mass of system)a 12 N = (2 + 4)kg (a), or 12 N/6 kg = 2 m/s2 = a

For part (b), let us stare at Fig. for #6b and c above. Let's cope with the massless rope first:

 (Fnet external)on rope = (mR)a Taking to the right to be positive, FRB - FRA = (0)a where a = the acceleration of the rope So, FRB = FRA

It is important to notice that FRB and FRA, while equal and opposite in direction, are not Newton third law of motion forces because they act on the SAME object. They are equal and in the opposite direction because they must sum to zero because a massless object has no net force acting on it. Then by Newton's third law we know that FRA = - FAR and FRB = -FBR. Thus since the magnitudes of these forces are equal, FRA = FRB, FAR = FBR. We throw all this careful notation to the wind and call FAR, FBR, FRA and FRB the tension T.

(b) Now we can choose either object to find the tension T in the rope.You can check the answer by isolating the other block and finding T.

 For block B, in the X-direction For block A, in the X-direction (Fnet external)x = (mB)a (Fnet external)x = (mA)a 12 N - T = (4 kg)(2 m/s2) T = (2 kg)(2 m/s2) = 4 N So 12 N - 8 N = T = 4 N 7.

The forces acting on the object are the applied force F and the weight of the object mg. In all cases, Fnet = ma (a) For constant velocity, a = 0, and F - mg = m(0) = 0, or F = mg = 3.0 kg (10 m/s2) = 30 N (b) For a constant acceleration, a = 3.0 m/s2 F - 30 N = 3.0 kg(3 m/s2) F = 9.0 N + 30 N = 39 N

8.

(a) For the entire system (Fig. 4a): Fnet = Ma F - 3mg = (3m)a 9.0 N - (0.30kg)(10 m/s2) = (0.30 kg)a 9.0 N - 3.0 N = (0.30 kg)a 20 m/s2 = a

(b) and (c) To find the tension at the top of the rope T(b), first use the lower                 figure in Fig. 4b below:

 Fnet = Ma T(b) - 2mg = (2m)a T(b) – 2(0.10 kg)(10 m/s2)     = (0.20 kg)(20m/s2) T(b) - 2.0 N = 4.0 N T(b) = 6.0 N Check with top figure in Fig. 4b. Fnet = Ma 9.0 N - mg - T(b) = ma 9.0 N – (0.10 kg)(10 m/s2) - T(b)     = (0.10kg)(20m/s2) = 2.0 N 6.0 N = T(b) (c)  The mass of 1/5 of the rope       = m/5 = 0.02 kg and its weight = mg/5 = 0.2N. (d) Use the top  figure in Fig. 4c. Fnet = Ma 9.0 N - 1.2 mg - T(c)     = (0.12 kg)(20 m/s2) 9.0 N – 1.2 N - T(c) = 2.4 N 7.8 N –2.4 N = Tc   or T(c) = 5.4 N Using the bottom figure in Fig. 4c Fnet = ma T(c) - 1.8 mg = (0.18 kg)(20 m/s2)     = 3.6 N T(c) = 5.4 N

9.

(a) For constant velocity, the acceleration is 0. From the graph of Fig. 5 below we see the applied force F = 1.0 N when the acceleration equals zero. The intercept on the applied axis is the frictional force f. In general, Fnet = ma. For the graph of Fig. 5,  F - f = ma, or  F = ma +f  (Equation 1).   If we compare Eq. 1 to the equation for a straight line, y = mx + b, we see that the intercept on the applied force F-axis is the frictional force f. (b) The slope of the graph is the mass of the object.
m = (4.0 - 1.0)N/(0.75 - 0)m/s2 = 4.0 kg.
Eq. 1 becomes F = 4.0 kg a + 1.0 N.

(c) For a = 1.0 m/s2, F = (4.0 kg)(1.0 m/s2) + 1.0 N or F = 5.0 N.

(d) For a block of a greater mass, the slope of the graph increases and the intercept also increases because the normal force N = mg and f = µN where µ is the coefficient of friction.

(e) If you change the surface, the intercept f changes because µ changes.

10.

 (a) (Fnet)x = max (Fnet)y = may 26 N - f = 2.0 kg(ax) FN - mg = m(0) FN =mg=(2.0 kg)(10m/s2) = 20N (b) f = µkFN = 1/5(20 N) = 4.0 N 26 N - 4.0 N = 2.0 kg(ax) (c) 11 m/s2 = ax

11.

 (a) (b) (Fnet)y = may (c) (Fnet)x = max FN + 26 N sin 22.60- mg = 0 26 N cos 22.60 - f = 2.0 kg(ax) FN + 26N(5/13) - 20 N = 0 26 N(12/13) - 2 N = 2.0 kg(ax) FN = 20 N - 26N(5/13)     = 20 N -10 N = 10N 24 N - 2 N = 2.0 kg(ax) f= µkFN = (1/5)(10 N) = 2 N 11 m/s2 = ax

12.

(a) and (b) The x-axis is chosen in the direction of the acceleration. The axes, forces and components of forces are shown in the figure below: (c) (Fnet)x = max 30 N sin 300 = 3.0 kg(ax) 15 N = 3.0 kg(ax) 5.0 m/s2 = ax

13. (b) Now there is a frictional force up the plane. (Fnet)y = may FN - mg cos 300 = 0 FN = 30 N (0.866) = 26 N f = µk FN = 0.154(26 N) = 4.0 N (c) (Fnet)x = max 30 N sin 300 - f = 3.0 kg(ax) 15 N - 4.0 N = 3.0 kg(ax) 11/3 m/s2 = ax

14. (a) First isolate m1, taking the X-axis in the direction of the acceleration.

 (Fnet)x = m1ax (Fnet)y = m1ay T - m1g sin 16.30 - f = m1ax FN - m1g cos16.30 = m1(0) FN = 25 N(24/25) = 24 N. f = µkFN = 1/6(24 N) = 4 N T - 25 N(7/25) - 4 N = 2.5 kg a T - 11 N = 2.5 kg a   (Equation #1)

I have dropped the subscript on a, the only acceleration. Now isolate m2, taking down as positive.

 Fnet = m2a m2g - T = m2a 20 N - T = 2.0 kg a   (Equation #2)

Adding Eq. 1 and Eq. 2:  20 N - 11 N = 4.5 kg a, or a = 2.0 m/s2

(b) Substituting a = 2.0 m/s2 into Eq. 2:
20 N - T = 2.0 kg(2.0 m/s2), and
T = 16 N
.

 15 Figure for #15a below is a sketch of the object moving on the table and all of the forces acting on it. As the object starts to slip away from the circle, a frictional force acts into the center of the circle to produce the centripetal acceleration. I chose the X-axis to the right because at the moment shown in the picture, the centripetal acceleration is into the center of the circle or to the right. The frictional force is the only force into the circle or, as I have drawn it, along the positive X-axis. This is a three dimensional drawing showing that for clockwise motion, at this instant of time, the velocity is along the positive Y-axis. The weight of the object is always along the negative Z-axis and the normal force of the table along the positive Z-axis. Now our axes of interests are the X-axis and the Z-axis. The acceleration in the positive X-direction is the centripetal acceleration = v2/r. (a) (Fnet external)x = max       Ffriction = mv2/r = 2 kg (2 m/s)2/(0.5 m) = 16 N (b) There is no acceleration in the Z-direction. (Fnet external)z = maz. FN - mg = m(0), or FN = mg = (2 kg)(10 m/s2) = 20 N. As shown in Fig. for #15b above: Force of table on object = {(Ffriction)2 + FN2)1/2 = {(16)2 + (20)2}1/2 = {656}1/2 N = 25.6 N. tan Θ = FN/Ffriction = 20/16 = 1.25; Θ = 51.30
 16 (a) Net force on sled = msledasled = 10 kg(2.0 m/s2) = 20 N = force of man on sled. (b) By Newton's third law, force of man on sled = force of sled on man = 20 N. (c) Net force on man = mmanaman = 60 kg(2.0 m/s2)= 120 N = frictional force of snow on man - force of sled on man 120 N = frictional force of snow on man - 20 N. Frictional force of snow on man = 140 N.

17. (a) The centripetal acceleration a = v2/r = (3m/s)2/0.5 m = 18 m/s2. The object travels in a counterclockwise direction so the directions of the velocities are as shown in Fig. 7a above. Remember the velocity is always tangent to the path. Thus,

(b) at B the velocity is to the right,

(c) at T the velocity is to the left, and

(d) at S the velocity is down.

(e) At B the acceleration is into the center of the circle or up. At B we take the X-axis up.

(f) At T the acceleration is into the circle or down. At T the X-axis is down.

(g) At S the acceleration is to the right and the X-axis is to the right. All of the accelerations are in the direction of the net force. For all cases Fnet = ma = 2 kg(18 m/s2) = 36 N. Looking at Fig. 7b below, we see that at B and T, there are no components of the force in any direction except the X-direction so we drop the subscripts on Fnet. (h) At B,
For at B - mg = ma
For at B - 20 N = 36 N
For at B = 56 N

(i) at T, I have drawn For at T down, but I must use the equation for a final decision so I write it with ± to find its sign and, therefore, its direction.
Fnet = ma ± For at T + 20 N = 36 N so For at T = +16 N or it is down.

(j) At S, For at S must have a component into the center or along the +X-axis to produce the centripetal acceleration and a component along the +Y-axis so the net force along the Y-axis is zero keeping the magnitude of the velocity for the uniform circular motion constant. Thus, we must look at Fnet along the X- and Y-axes.

 (Fnet)x at S = max (Fnet)y at S = may (For)x at S = 36 N (For)y at S - mg = m(0) (For)y at S = mg = 20 N For at S = {(For)2x at S + (For)2x at S} For at S = {(36)2 + (20)2}1/2 N = 41 N tan Θ = (For)y/(For)x = 20 N/ 36 N = 0.55. Θ = 290

18.

(a) Let Fos = the force of the scale on the object when the elevator moves with a constant velocity. For constant velocity, Fnet = ma = m(0) Fos - mg = 0, and Fos = mg = (10 kg)(10 m/s2) =100 N

By Newton's third law, the force of the scale on the object = the force of the object on the scale. The reading of the scale is 100 N.

(b) Let F'os = the force of the scale on the object when the elevator moves upward with a constant acceleration of 5 m/s2.

 Fnet = ma = (10 kg)(5 m/s2) F'os - mg = 50 N, or F'os = 50 N + 100 N = 150 N F'os = F'so = reading of scale= 150 N

 19 From problem #18, we see that the scale reading is greater for an upward acceleration. Conversely the scale reading is smaller for a downward acceleration. We can explain Fig. 8 below by saying that at first the elevator accelerates upward, then moves at a constant velocity, and finally decelerates and comes to rest. 20 Fnet = mg - f = ma or a = g - f/m. The lead sphere has a greater density than the wood sphere. The volume of the spheres = 4/3 πr3 and the mass m of the spheres = density x volume. The lead sphere has the same volume as the wood sphere since their radii are identical, but it has a larger mass because it has a greater density. Since a = g - f/m, the sphere of the larger mass has the greater acceleration.

21.

Take the X-axis in the direction of the acceleration or up the plane and the Y-axis perpendicular to it (Fig. for #21). (Fnet)x = max (Fnet)y = may F cos 370 - mg sin 370 - f = ma FN - F sin 370 -mg cos 370 = m(0) 60 N(4/5) - 20 N(3/5) - f = 2.0 kg(a) FN = 60 N(3/5) + 20 N(4/5) = 52 N f = µkFN = 1/2(52 N) = 26 N 48 N - 12 N - 26 N = 2.0 kg(a) or a = 5.0 m/s2

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