Ch5_ProblemsSet - Chapter 5 Problems Set 5 Three astronauts propelled by jet backpacks push and guide a 120 kg asteroid toward a processing 0 dock

# Ch5_ProblemsSet - Chapter 5 Problems Set 5 Three astronauts...

This preview shows page 1 out of 5 pages.

#### You've reached the end of your free preview.

Want to read all 5 pages?

Unformatted text preview: Chapter 5 – Problems Set 5. Three astronauts, propelled by jet backpacks, push and guide a 120 kg asteroid toward a processing 0 dock, exerting the forces shown in Fig. 5-­‐29 (see book), with F1=32 N, F2=55 N, F3=41 N, θ1=30 , and 0 θ3=60 . What is the asteroid's acceleration (a) in unit-­‐vector notation and as (b) a magnitude and (c) a direction relative to the positive direction of the x axis? 11. A 2.0 kg particle moves along an x axis, being propelled by a variable force directed along that axis. Its position is given by x = 3.0m + (4.0m / s)t + ct − (2.0m / s )t , with x in meters and t in seconds. The factor c is a constant. At t=3.0s, the force on the particle has a magnitude of 36 N and is in the negative direction of the axis. What is c? 2 3 3 13. Figure 5-­‐33 (See book) shows an arrangement in which four disks are suspended by cords. The longer, top cord loops over a frictionless pulley and pulls with a force of magnitude 98 N on the wall to which it is attached. The tensions in the three shorter cords are T1=58.8 N, T2=49.0 N, and T3=9.8 N. What are the masses of (a) disk A, (b) disk B, (c) disk C, and (d) disk D? 27. An electron with a speed of 1.2x107 m/s moves horizontally into a region where a constant vertical force of 4.5x10-­‐16 N acts on it. The mass of the electron is 9.11x10-­‐31 kg. Determine the vertical distance the electron is deflected during the time it has moved 30 mm horizontally. 34. In Fig. 5-­‐40 (see book), a crate of mass m=100 kg is pushed at constant speed up a frictionless ramp ! ! (θ=30.00) by a horizontal force F . What are the magnitudes of (a) F and (b) the force on the crate from the ramp? 50. In Fig. 5-­‐46 (see book), three ballot boxes are connected by cords, one of which wraps over a pulley having negligible friction on its axle and negligible mass. The three masses are mA=30.0 kg, mB=40.0 kg, and mC=10.0 kg. When the assembly is released from rest, (a) what is the tension in the cord connecting B and C, and (b) how far does A move in the first 0.250 s (assuming it does not reach the pulley)? 53. In Fig. 5-­‐48 (see book), three connected blocks are pulled to the right on a horizontal frictionless table by a force of magnitude T3=65.0 N. If m1=12.0 kg, m2=24.0 kg, and m3=31.0 kg, calculate (a) the magnitude of the system's acceleration, (b) the tension T1, and (c) the tension T2. 57. A block of mass m1=3.70 kg on a frictionless plane inclined at angle θ=30.00 is connected by a cord over a massless, frictionless pulley to a second block of mass m2=2.30 kg (Fig. 5-­‐52 – see book). What are (a) the magnitude of the acceleration of each block, (b) the direction of the acceleration of the hanging block, and (c) the tension in the cord? 61. A hot-­‐air balloon of mass M is descending vertically with downward acceleration of magnitude a. How much mass (ballast) must be thrown out to give the balloon an upward acceleration of magnitude a? Assume that the upward force from the air (the lift) does not change because of the decrease in mass. ! 78. In Fig. 5-­‐64 (see book), a force F of magnitude 12 N is applied to a FedEx box of mass m2=1.0 kg. The force is directed up a plane tilted by θ=37.00. The box is connected by a cord to a UPS box of mass m1=3.0 kg on the floor. The floor, plane, and pulley are frictionless, and the masses of the pulley and cord are negligible. What is the tension in the cord? 5. The net force applied on the chopping block is , where the vector addition is done using unit-­‐vector notation. (a) The forces exerted by the three astronauts can be expressed in unit-­‐vector notation as follows: The resultant acceleration of the asteroid of mass m = 120 kg is therefore (b) The magnitude of the acceleration vector is (c) The vector makes an angle q with the +x axis, where 11. The velocity is the derivative (with respect to time) of given function x, and the acceleration is the derivative of the velocity. Thus, a = 2c – 3(2.0)(2.0)t, which we use in Newton’s second law: F = (2.0 kg)a = 4.0c – 24t (with SI units understood). At t = 3.0 s, we are told that F = –36 N. Thus, –36 = 4.0c – 24(3.0) can be used to solve for c. The result is c = +9.0 m/s2. 13. (a) From the fact that T3 = 9.8 N, we conclude the mass of disk D is 1.0 kg. Both this and that of disk C cause the tension T2 = 49 N, which allows us to conclude that disk C has a mass of 4.0 kg. The weights of these two disks plus that of disk B determine the tension T1 = 58.8 N, which leads to the conclusion that mB = 1.0 kg. The weights of all the disks must add to the 98 N force described in the problem; therefore, disk A has mass 4.0 kg. (b) mB = 1.0 kg, as found in part (a). (c) mC = 4.0 kg, as found in part (a). (d) mD = 1.0 kg, as found in part (a). 27. THINK An electron moving horizontally is under the influence of a vertical force. Its path will be deflected toward the direction of the applied force. EXPRESS The setup is shown in the figure below. The acceleration of the electron is vertical and for all practical purposes the only force acting on it is the electric force. The force of gravity is negligible. We take the +x axis to be in the direction of the initial velocity v0 and the +y axis to be in the direction of the electrical force, and place the origin at the initial position of the electron. Since the force and acceleration are constant, we use the equations from Table 2-­‐1: and ANALYZE The time taken by the electron to travel a distance x (= 30 mm) horizontally is t = x/v0 and its deflection in the direction of the force is LEARN Since the applied force is constant, the acceleration in the y-­‐direction is also constant and the path is parabolic with 34. We resolve this horizontal force into appropriate components. (a) Newton’s second law applied to the x-­‐axis produces For a = 0, this yields F = 566 N. (b) Applying Newton’s second law to the y axis (where there is no acceleration), we have which yields the normal force FN = 1.13 ´ 103 N. 50. (a) The net force on the system (of total mass M = 80.0 kg) is the force of gravity acting on the total overhanging mass (mBC = 50.0 kg). The magnitude of the acceleration is therefore a = (mBC g)/M = 6.125 m/s2. Next we apply Newton’s second law to block C itself (choosing down as the +y direction) and obtain mC g – TBC = mC a. This leads to TBC = 36.8 N. 1 (b) We use Eq. 2-­‐15 (choosing rightward as the +x direction): Dx = 0 + 2 at2 = 0.191 m. 53. We apply Newton’s second law first to the three blocks as a single system and then to the individual blocks. The +x direction is to the right in Fig. 5-­‐48. (a) With msys = m1 + m2 + m3 = 67.0 kg, we apply Eq. 5-­‐2 to the x motion of the system, in which case, there is only one force . Therefore, 2 which yields a = 0.970 m/s for the system (and for each of the blocks individually). (b) Applying Eq. 5-­‐2 to block 1, we find (c) In order to find T2, we can either analyze the forces on block 3 or we can treat blocks 1 and 2 as a system and examine its forces. We choose the latter. 57. The free-­‐body diagram for each block is shown below. T is the tension in the cord and q = 30° is the angle of the incline. For block 1, we take the +x direction to be up the incline and the +y direction to be in the direction of the normal force that the plane exerts on the block. For block 2, we take the +y direction to be down. In this way, the accelerations of the two blocks can be represented by the same symbol a, without ambiguity. Applying Newton’s second law to the x and y axes for block 1 and to the y axis of block 2, we obtain respectively. The first and third of these equations provide a simultaneous set for obtaining values of a and T. The second equation is not needed in this problem, since the normal force is neither asked for nor is it needed as part of some further computation (such as can occur in formulas for friction). (a) We add the first and third equations above: m2g – m1g sin q = m1a + m2a. Consequently, we find (b) The result for a is positive, indicating that the acceleration of block 1 is indeed up the incline and that the acceleration of block 2 is vertically down. (c) The tension in the cord is 61. THINK As more mass is thrown out of the hot-­‐air balloon, its upward acceleration increases. EXPRESS The forces on the balloon are the force of gravity (down) and the force of the air (up). We take the +y to be up, and use a to mean the magnitude of the acceleration. When the mass is M (before the ballast is thrown out) the acceleration is downward and Newton’s second law is After the ballast is thrown out, the mass is M – m (where m is the mass of the ballast) and the acceleration is now upward. Newton’s second law leads to Fa – (M – m)g = (M – m)a. Combing the two equations allows us to solve for m. ANALYZE The first equation gives Fa = M(g – a), and this plugs into the new equation to give M ( g − a) − ( M − m) g = ( M − m) a ⇒ m = LEARN More generally, if a ballast mass via: 2Ma . g+a is tossed, the resulting acceleration is which is related to , showing that the more mass thrown out, the greater is the upward acceleration. For , which agrees with what was found above. , we get ...
View Full Document

• One '14