Question

A 2.50-kg object is moving in a plane, with its x and y coordinates given by x = 3t² − 4 and y = 2t³ + 5, where x and y are in meters and t is in seconds. Find the magnitude of the net force acting on this object at t = 1.90 s.

Answer 1

The magnitude of the net force acting on the object at t = 1.90 s is approximately 33.54 N.

Step-by-step explanation:

To find the magnitude of the net force acting on the object at t = 1.90 s, we first need to find the x and y components of the force. The net force can be calculated using the formula: Fnet = (m * ax)î + (m * ay)ĵ. In this case, the mass (m) of the object is 2.50 kg. To find the accelerations ax and ay, we need to differentiate the given equations for x and y with respect to t.

The x-component of the net force:

ax = d²x/dt² = 2 * 3 = 6 m/s²

The y-component of the net force:

ay = d²y/dt² = 2 * 3 * 2 = 12 m/s²

Now, we can calculate the magnitude of the net force using the formula:

|Fnet| = sqrt(Fx² + Fy²)

|Fnet| = sqrt((m * ax)² + (m * ay)²)

|Fnet| = sqrt((2.50 * 6)² + (2.50 * 12)²)

|Fnet| = sqrt(225 + 900)

|Fnet| = sqrt(1125)

|Fnet| ≈ 33.54 N