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A 12.0 N force with a fixed orientation performs work on a particle as it moves through a three-dimensional displacement m. What is the angle between the force and the displacement if the change in the particle's kinetic energy is (a) 30.0 J and (b) 30.0 J?

User Nick Baluk
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Final answer:

The angle between the force and displacement cannot be calculated without the value of the displacement. However, if the full 12.0 N force results in 30.0 J of kinetic energy, the angle is likely zero degrees, since the cosine of zero is 1, and any other angle would require a greater displacement.

Step-by-step explanation:

To determine the angle between the force and the particle's displacement when a change in kinetic energy occurs, we can use the work-energy theorem. The work-energy theorem states that the work done on an object is equal to its change in kinetic energy (ΔK). The work done by a force through a displacement is given by the equation W = F * d * cos(θ), where W is work, F is the magnitude of the force, d is the displacement, and θ is the angle between the force and displacement.

In this case, the force is 12.0 N, and the change in kinetic energy given is 30.0 J. Assuming the displacement is in the direction of the force, we set up the equation:

30.0 J = 12.0 N * d * cos(θ)

Since the displacement is not given, we continue under the assumption that it is in the direction of the force, which would imply θ = 0. Because the change in kinetic energy is positive, we know the work is also positive, indicating that the force is doing work in the direction of the displacement. Thus, θ must be either 0 degrees or an acute angle, as cos(θ) > 0 in these cases.

Without the value of 'm', the displacement, we can't solve for the angle mathematically. However, we can infer that if the full 12.0 N force is contributing to the 30.0 J of kinetic energy, then the cosine term must equal 1, which suggests a zero-degree angle between the force and the displacement. For any non-zero angle, the displacement 'm' would need to be even greater to maintain the 30.0 J of work being performed.

User TrN
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