In sample Prob. 9.03, if a forth particle ofmass 1 kg is experiencing a force ofmagnitude F4 = -2j (in negative y direction).(a) What is the new acceleration of thecenter of mass of the system? (b) in whatdirection does it move?O a. (a) a(com) = 1.2 m/s^2, (b) theta =%3D63 with the +x axisb. (a) a(com) = 1.2 m/s^2, (b) theta =27 degrees with the +x axis%3Dc. a) a(com) = 1.1 m/s^2, (b) theta%D70 degrees with the +x axisd. (a) a(com) = 1.1 m/s^2, (b) theta%3D20 degrees with the +x ais Sample Problem 9.03 Motion of the com of three particlesIf the particles in a system all move together, the com moveswith them-no trouble there. But what happens when theymove in different directions with different accelerations?Here is an example.The three particles in Fig. 9-7a are initially at rest. Eachexperiences an external force due to bodies outside thethree-particle system. The directions are indicated, and themagnitudes are F = 6.0 N, F, = 12 N, and F = 14 N. Whatis the acceleration of the center of mass of the system, and inwhat direction does it move?Along the y axis, we haveFy + F + Fa com, y0 + (12 N) sin 45° + 00.530 m/s.16 kgFrom these components, we find that acom has the magnitudea comV(a com.) + (acom.y)= 1.16 m/s? 1.2 m/sand the angle (from the positive direction of the x axis)(Answer)KEY IDEASA com.y-10 = tan= 27°.(Answer)The position of the center of mass is marked by a dot in thefigure. We can treat the center of mass as if it were a realparticle, with a mass equal to the system's total mass M = 16 kg.We can also treat the three external forces as if they act at thecenter of mass (Fig. 9-7b).a com, x4.0 kg45°Calculations: We can now apply Newton's second law8.0 kg(Fnet = md) to the center of mass, writingcomoFnet= Mac(9-20)com-3-2-124F+ F, + F, = Ma comor-1F+ F, + F4.0 kg-2d com(9-21)soThe com of the system3will move as if all the(a)Equation 9-20 tells us that the acceleration dcom of thecenter of mass is in the same direction as the net external forcemass were there andthe net force acted there.InetFnet on the system (Fig. 9-7b). Because the particles are ini-tially at rest, the center of mass must also be at rest. As thecenter of mass then begins to accelerate, it must move off inthe common direction of aWe can evaluate the right side of Eq. 9-21 directly ona vector-capable calculator, or we can rewrite Eq. 9-21 incomponent form, find the components of aom, and then findaM= 16 kge "comand F-netcomcomF.10-3 -2-12.3.4(b)Along the x axis, we haveFigure 9-7 (a) Three particles, initially at rest in the positions shown,are acted on by the external forces shown. The center of mass (com)of the system is marked. (b) The forces are now transferred to thecenter of mass of the system, which behaves like a particle with amass M equal to the total mass of the system. The net external forceF and the acceleration aom of the center of mass are shown.comF, + F + Fa com, x-6.0 N + (12 N) cos 45° + 14 N= 1.03 m/s2.16 kgnetractice available at WileyPLUS

Question

In sample Prob. 9.03, if a forth particle ofmass 1 kg is experiencing a force ofmagnitude F4 = -2j (in negative y direction).(a) What is the new acceleration of thecenter of mass of the system? (b) in whatdirection does it move?O a. (a) a(com) = 1.2 m/s^2, (b) theta =%3D63 with the +x axisb. (a) a(com) = 1.2 m/s^2, (b) theta =27 degrees with the +x axis%3Dc. a) a(com) = 1.1 m/s^2, (b) theta%D70 degrees with the +x axisd. (a) a(com) = 1.1 m/s^2, (b) theta%3D20 degrees with the +x ais Sample Problem 9.03 Motion of the com of three particlesIf the particles in a system all move together, the com moveswith them-no trouble there. But what happens when theymove in different directions with different accelerations?Here is an example.The three particles in Fig. 9-7a are initially at rest. Eachexperiences an external force due to bodies outside thethree-particle system. The directions are indicated, and themagnitudes are F = 6.0 N, F, = 12 N, and F = 14 N. Whatis the acceleration of the center of mass of the system, and inwhat direction does it move?Along the y axis, we haveFy + F + Fa com, y0 + (12 N) sin 45° + 00.530 m/s.16 kgFrom these components, we find that acom has the magnitudea comV(a com.) + (acom.y)= 1.16 m/s? 1.2 m/sand the angle (from the positive direction of the x axis)(Answer)KEY IDEASA com.y-10 = tan= 27°.(Answer)The position of the center of mass is marked by a dot in thefigure. We can treat the center of mass as if it were a realparticle, with a mass equal to the system's total mass M = 16 kg.We can also treat the three external forces as if they act at thecenter of mass (Fig. 9-7b).a com, x4.0 kg45°Calculations: We can now apply Newton's second law8.0 kg(Fnet = md) to the center of mass, writingcomoFnet= Mac(9-20)com-3-2-124F+ F, + F, = Ma comor-1F+ F, + F4.0 kg-2d com(9-21)soThe com of the system3will move as if all the(a)Equation 9-20 tells us that the acceleration dcom of thecenter of mass is in the same direction as the net external forcemass were there andthe net force acted there.InetFnet on the system (Fig. 9-7b). Because the particles are ini-tially at rest, the center of mass must also be at rest. As thecenter of mass then begins to accelerate, it must move off inthe common direction of aWe can evaluate the right side of Eq. 9-21 directly ona vector-capable calculator, or we can rewrite Eq. 9-21 incomponent form, find the components of aom, and then findaM= 16 kge &#34;comand F-netcomcomF.10-3 -2-12.3.4(b)Along the x axis, we haveFigure 9-7 (a) Three particles, initially at rest in the positions shown,are acted on by the external forces shown. The center of mass (com)of the system is marked. (b) The forces are now transferred to thecenter of mass of the system, which behaves like a particle with amass M equal to the total mass of the system. The net external forceF and the acceleration aom of the center of mass are shown.comF, + F + Fa com, x-6.0 N + (12 N) cos 45° + 14 N= 1.03 m/s2.16 kgnetractice available at WileyPLUS

Accepted Answer

Given data:

The fourth force is F4=-2 N.
Mass of fourth particle is m=1 kg.