A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, 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 = 2.15 s.

Explanation:

The sum of force along both components x and y is expressed as;

The magnitude of the net force which is also known as the resultant will be expressed as

To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

Given the position of the object along the x-component to be x = 6t² − 4;

Similarly,

Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N

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