215k views
2 votes
A 3.0-kg object is moving in a plane, with its x and y coordinates given by x = 5t^2 – 1 m and y = 3t^3 + 2 m. Find the magnitude and direction of the net force acting on the object.

1 Answer

2 votes

Final answer:

The magnitude of the net force on the object varies with respect to time due to the variable acceleration in the y-direction. It is computed using the Pythagorean theorem applied to the forces in the x and y directions, while the direction is obtained by taking the inverse tangent of the forces' ratio.

Step-by-step explanation:

To find the magnitude and direction of the net force on the 3.0-kg object, we first need to determine its acceleration in both the x and y directions. The acceleration is the derivative of the velocity, which in turn is the derivative of the position. Given the equations for x and y, we can differentiate x = 5t^2 – 1 m to get derivative dx/dt = 10t m/s, which is the velocity in the x direction. Furthermore, differentiating the velocity gives us derivative dvx/dt = 10 m/s^2, which is the acceleration in the x direction. Similarly, for y = 3t^3 + 2 m, dy/dt = 9t^2 m/s (velocity in y), and derivative dvy/dt = 18t m/s^2 (acceleration in y).

Now, we can compute the net force using Newton's second law, F=ma. In the x direction, Fx = m * ax = 3 kg * 10 m/s^2 = 30 N. In the y direction, Fy = m * ay = 3 kg * 18t m/s^2 = 54t N, with t being the variable with respect to time.

The magnitude of the net force is the square root of the sum of the squares of the forces in each direction, √(Fx^2 + Fy^2) = √((30 N)^2 + (54t N)^2), and the direction is found using the inverse tangent of the ratio of the forces (θ= tan^-1 (Fy/Fx)).

Learn more about Net Force

User Jason Nordwick
by
8.5k points

No related questions found