Question

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 object 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.

Answer 1

The work done by the force is 254.0 Joules.

Step-by-step explanation:

In order to find the work done by the force, we need to use the formula:

W = F · d

where W is the work, F is the force, and d is the displacement. The dot product of the force and displacement vectors will give us the work done.

In this case, the force vector is (16.0 i^ + 29.0 j^) N and the displacement vector can be found by subtracting the initial position vector from the final position vector. The work done is the dot product of these vectors.

W = (16.0 i^ + 29.0 j^) N · (5.0 i^ + 6.0 j^) m

= (16.0 * 5.0) Nm + (29.0 * 6.0) Nm

= 80.0 Nm + 174.0 Nm

= 254.0 Nm

Therefore, the work done by the force is 254.0 Joules.

Answer 2

The work done by the force is 820.745 joules.

Step-by-step 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) = (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\,(m)/(s) \right)^(2)+\left(22\,(m)/(s) \right)^(2)}`

`v_(1) \approx 25.060\,(m)/(s)`

Final velocity

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

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

`v_(2) \approx 33.121\,(m)/(s)`

Finally, if
`m = 3.5\,kg`,
`v_(1) \approx 25.060\,(m)/(s)` and
`v_(2) \approx 33.121\,(m)/(s)`, then the work done by the force is:

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

`W_(F) = 820.745\,J`

The work done by the force is 820.745 joules.