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A force of fx = (10N)sin(2πt/4.0s) is exerted on a 320g particle during the interval 0s ≤ t ≤ 2.0s. What is the complete question?

User WilliamLou
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Final answer:

To find the work done by the given force, we need to calculate the area under the force-time graph.

Step-by-step explanation:

The given force can be represented as fx = 10N*sin(2πt/4.0s). This force is exerted on a particle with a mass of 320g from 0s to 2.0s.

To find the work done by this force, we can use the formula for work: Work = Force * Displacement * cos(theta). Since the force is in the x-direction and there is no displacement in the y-direction, the angle theta is 0 degrees.

Thus, the work done by the force is equal to the area under the force-time graph, which is the integral of the force over the given time interval.

User Benni
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