Question

A 2.65-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 2 and y = 4t3 + 3, 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 = 1.65 s.

N

Answer 1

Step-by-step explanation:

To find the magnitude of the net force acting on the object, we need to find the acceleration of the object at t = 1.65 s.

The acceleration can be found by taking the second derivative of the position function with respect to time:

x = 6t^2 - 2

y = 4t^3 + 3

Taking the second derivative of x with respect to t:

d^2x/dt^2 = 12

Taking the second derivative of y with respect to t:

d^2y/dt^2 = 24t

Now, we can find the acceleration at t = 1.65 s:

a = sqrt((d^2x/dt^2)^2 + (d^2y/dt^2)^2)

a = sqrt((12)^2 + (24(1.65))^2)

a = sqrt(144 + 957.6)

a ≈ sqrt(1101.6)

a ≈ 33.18 m/s^2

Finally, we can calculate the magnitude of the net force using Newton's second law, F = ma:

F = ma

F = (2.65 kg)(33.18 m/s^2)

F ≈ 87.90 N

Therefore, the magnitude of the net force acting on the object at t = 1.65 s is approximately 87.90 N.