A 2.50-kg object is moving along the x -axis at 1.60 m/s. As it S S passes the origin, two forces F 1 and F 2 are applied, both in the y -direction (plus or minus). The forces are applied for 3.00 s, after S which the object is at x S = 4.80 m, y = 10.8 m. If F 1 = 15.0 N, what’s F 2 ?

Question

A 2.50-kg object is moving along the x -axis at 1.60 m/s. As it S S passes the origin, two forces F 1 and F 2 are applied, both in the y -direction (plus or minus). The forces are applied for 3.00 s, after S which the object is at x S = 4.80 m, y = 10.8 m. If F 1 = 15.0 N, what’s F 2 ?

Answer 1

Just do the y problem. Motion or force in the x direction has nothing to do with it. y momentum at t = 3 is y force * time

Answer 2

To find the value of F2, we need to use the given information and apply the principles of Newton's laws of motion.

First, let's break down the problem and identify the known values:

Mass of the object (m) = 2.50 kg
Initial velocity along the x-axis (v) = 1.60 m/s

Force F1 = 15.0 N

Time (t) = 3.00 s

Final position (xS, y) = (4.80 m, 10.8 m)

We can start by analyzing the motion of the object in the x-axis. Since no force is applied in the x-direction, the object will keep moving with a constant velocity. We can find the displacement in the x-direction using the initial velocity and time:

Δx = v * t
Δx = 1.60 m/s * 3.00 s
Δx = 4.80 m

So the object moved 4.80 m in the x-direction during the 3.00 s time interval.

Now, let's consider the motion in the y-direction. Since forces F1 and F2 are applied in the y-direction, they will cause the object to accelerate. Using Newton's second law of motion, we can determine the net force acting on the object in the y-direction.

Fnet = ma

In the y-direction, the only forces acting are F1 and F2, so we can rewrite the equation as:

F1 + F2 = ma

Since we know the mass of the object is 2.50 kg, we can substitute this value into the equation:

15.0 N + F2 = 2.50 kg * a ---- (Equation 1)

Next, we need to find the acceleration (a). We can use the y-component of the final position and the initial velocity to find the displacement in the y-direction:

Δy = v * t + (1/2) * a * t^2
10.8 m = 0 + (1/2) * a * (3.00 s)^2
10.8 m = (1.5) * a * 9.00 s^2
10.8 m = 13.5 a

Now, we can solve for a:

a = 10.8 m / 13.5
a = 0.8 m/s^2

Substituting this value for a in Equation 1, we get:

15.0 N + F2 = 2.50 kg * 0.8 m/s^2
15.0 N + F2 = 2.00 N

Finally, solving for F2:

F2 = 2.00 N - 15.0 N
F2 = -13.0 N

Therefore, the value of F2 is -13.0 N. The negative sign indicates that the force F2 is acting in the opposite direction of F1.