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Vesna [10]
2 years ago
11

A 3.5 kg object moving in two dimensions initially has a velocity v1 = (12.0 i^ + 22.0 j^) m/s. A net force F then acts on the o

bject for 2.0 s, after which the object's velocity is v 2 = (16.0 i^ + 29.0 j^) m/s.
Required:
Find the work done by the force in joules.
Physics
1 answer:
lys-0071 [83]2 years ago
6 0

Answer:

The work done by the force is 820.745 joules.

Explanation:

Let suppose that changes in potential energy can be neglected. According to the Work-Energy Theorem, an external conservative force generates a change in the state of motion of the object, that is a change in kinetic energy. This phenomenon is describe by the following mathematical model:

K_{1} + W_{F} = K_{2}

Where:

W_{F} - Work done by the external force, measured in joules.

K_{1}, K_{2} - Translational potential energy, measured in joules.

The work done by the external force is now cleared within:

W_{F} = K_{2} - K_{1}

After using the definition of translational kinetic energy, the previous expression is now expanded as a function of mass and initial and final speeds of the object:

W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})

Where:

m - Mass of the object, measured in kilograms.

v_{1}, v_{2} - Initial and final speeds of the object, measured in meters per second.

Now, each speed is the magnitude of respective velocity vector:

Initial velocity

v_{1} = \sqrt{v_{1,x}^{2}+v_{1,y}^{2}}

v_{1} = \sqrt{\left(12\,\frac{m}{s} \right)^{2}+\left(22\,\frac{m}{s} \right)^{2}}

v_{1} \approx 25.060\,\frac{m}{s}

Final velocity

v_{2} = \sqrt{v_{2,x}^{2}+v_{2,y}^{2}}

v_{2} = \sqrt{\left(16\,\frac{m}{s} \right)^{2}+\left(29\,\frac{m}{s} \right)^{2}}

v_{2} \approx 33.121\,\frac{m}{s}

Finally, if m = 3.5\,kg, v_{1} \approx 25.060\,\frac{m}{s} and v_{2} \approx 33.121\,\frac{m}{s}, then the work done by the force is:

W_{F} = \frac{1}{2}\cdot (3.5\,kg)\cdot \left[\left(33.121\,\frac{m}{s} \right)^{2}-\left(25.060\,\frac{m}{s} \right)^{2}\right]

W_{F} = 820.745\,J

The work done by the force is 820.745 joules.

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Explanation:

Position of centre of mass is given as

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