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As a 4.2-kg object moves from (2 i + 5 j) m to (6 i - 2 j) m, the constant resultant force acting on it is equal to (4 i - 3 j) N. If the speed of the object at the initial position is 2.5 m/s, what is the work done by the force, and what is its kinetic energy at its final position?

User Kafels
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Final answer:

The work done by the force is 37 J, and the final kinetic energy of the object is 50.125 J after solving the problems using the dot product for work done and the work-energy theorem for kinetic energy.

Step-by-step explanation:

The work done by a constant force when an object moves through a displacement can be found using the formula Work = Force × Displacement × cos(ϴ), where ϴ is the angle between the force and displacement vectors. Since the force here is constant and given as (4i - 3j) N, we can calculate the displacement vector as the final position vector minus the initial position vector, which is (6i - 2j) m - (2i + 5j) m, resulting in (4i - 7j) m. The work done is then the dot product of force and displacement vectors, Work = (4 × 4) + (-3 × -7) = 16 + 21 = 37 J. To find the kinetic energy at the final position, we use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. The initial kinetic energy is given by ½mv², where m is the mass of the object and v is its velocity, so ½ × 4.2 kg × (2.5 m/s)² = 13.125 J. Since the work done on the object is positive, it adds to the initial kinetic energy, yielding a final kinetic energy of 13.125 J + 37 J = 50.125 J.

User Bbunmp
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