The location of two particles at t = 0 is shown in the figure below. The particles are initially at rest. Each particle is subject to a constant force, as indicated in the figure. (Take m1 = 2.06 kg and m2 = 0.95 kg.)

(a) Find the position of the center of mass at t = 0.
xcm = 1 cm
ycm = cm

(b) Find the resultant force acting on the system.
magnitude N
direction ° above the x-axis

(c) Find the location of the center of mass at t = 5.15 s.
xcm = m
ycm = m

To find the position of the center of mass at t = 0, we need to calculate the x-coordinate and the y-coordinate separately.

(a) The x-coordinate of the center of mass (xcm) is given by the formula:
xcm = (m1 * x1 + m2 * x2) / (m1 + m2)
Where m1 and x1 are the mass and x-coordinate of the first particle, and m2 and x2 are the mass and x-coordinate of the second particle.

In the given question, the mass of the first particle (m1) is 2.06 kg and its x-coordinate is 2 cm (from the figure). The mass of the second particle (m2) is 0.95 kg and its x-coordinate is -1 cm (from the figure).
Using the formula, we can calculate the x-coordinate of the center of mass:

xcm = (2.06 kg * 2 cm + 0.95 kg * -1 cm) / (2.06 kg + 0.95 kg)
xcm = (4.12 cm - 0.95 cm) / 2.25 kg
xcm = 3.17 cm / 2.25 kg
xcm = 1.41 cm

Therefore, the x-coordinate of the center of mass at t = 0 is 1.41 cm.

For the y-coordinate of the center of mass (ycm), we can see from the figure that both particles are initially at rest along the y-axis. Hence, the y-coordinate of the center of mass will be the same as the y-coordinate of the particles.

(b) To find the resultant force acting on the system, we can add the forces acting on each particle together. The resultant force (F_res) is the vector sum of the individual forces. We can find the magnitude and direction of the resultant force.

Magnitude of F_res:
Using the Pythagorean theorem, we can find the magnitude of the resultant force:
F_res = √(Fx^2 + Fy^2)
Where Fx is the sum of the x-components of the forces, and Fy is the sum of the y-components of the forces.

From the figure, we can see that the x-components of the forces are -5 N and 3 N, and the y-components of the forces are -4 N and -6 N. Plugging these values into the formula, we get:
F_res = √((-5 N + 3 N)^2 + (-4 N + -6 N)^2)
F_res = √((-2 N)^2 + (-10 N)^2)
F_res = √(4 N^2 + 100 N^2)
F_res = √(104 N^2)
F_res = √(104) N
F_res ≈ 10.2 N (rounded to one decimal place)

Direction of F_res:
To find the direction of the resultant force, we can use trigonometry. The angle (θ) can be calculated using the inverse tangent (arctan) function:
θ = arctan(Fy / Fx)

Plugging the values, we get:
θ = arctan((-4 N + -6 N) / (-5 N + 3 N))
θ = arctan((-10 N) / (-2 N))
θ = arctan(5)

Using a calculator, we find:
θ ≈ 78.69° (rounded to two decimal places)

Therefore, the magnitude of the resultant force is approximately 10.2 N, and its direction is approximately 78.69° above the x-axis.

(c) To find the location of the center of mass at t = 5.15 s, we need to use the equations of motion. Since the question does not provide any information about the motion of the particles, we cannot determine the exact positions at a specific time without additional information.