1.   A force F = 2i - yj, with x in meters, that acts on a particle as it moves from coordinates (-1,5) to (2,-4). What is the work done?

To calculate the work done by a force on a particle, we use the formula:

\[
W = \int_{C} \mathbf{F} \cdot d\mathbf{r}
\]

Where:
- \( \mathbf{F} = 2i - yj \) is the force acting on the particle.
- \( d\mathbf{r} = dx \, i + dy \, j \) is the differential displacement.
- The path \( C \) is from \( (-1, 5) \) to \( (2, -4) \).

We can parametrize the path of the particle as \( \mathbf{r}(t) = x(t) \, i + y(t) \, j \), where \( t \in [0, 1] \). We assume a linear path between the two points, meaning:

\[
x(t) = -1 + 3t, \quad y(t) = 5 - 9t
\]

Thus, the velocity or displacement differential is:

\[
d\mathbf{r} = \frac{dx}{dt} \, dt \, i + \frac{dy}{dt} \, dt \, j = 3i - 9j \, dt
\]

Now, the force is:

\[
\mathbf{F}(x, y) = 2i - yj
\]

Substitute the parametric expressions for \( x(t) \) and \( y(t) \) into the force:

\[
\mathbf{F}(t) = 2i - (5 - 9t)j
\]

Next, calculate the dot product \( \mathbf{F}(t) \cdot d\mathbf{r} \):

\[
\mathbf{F}(t) \cdot d\mathbf{r} = (2i - (5 - 9t)j) \cdot (3i - 9j) \, dt
\]

Performing the dot product:

\[
\mathbf{F}(t) \cdot d\mathbf{r} = (2 \cdot 3) + (- (5 - 9t) \cdot (-9)) \, dt
= 6 + 9(5 - 9t) \, dt
\]
\[
= 6 + 45 - 81t \, dt = (51 - 81t) \, dt
\]

Finally, we integrate this expression from \( t = 0 \) to \( t = 1 \):

\[
W = \int_0^1 (51 - 81t) \, dt = \left[ 51t - \frac{81t^2}{2} \right]_0^1
\]
\[
= \left( 51(1) - \frac{81(1)^2}{2} \right) - \left( 51(0) - \frac{81(0)^2}{2} \right)
\]
\[
= 51 - \frac{81}{2} = 51 - 40.5 = 10.5 \, \text{J}
\]

The work done by the force is \( 10.5 \, \text{J} \).

Would you like further clarification, or details about any steps?

Here are 5 related questions for you:
1. What is the general relationship between force and work?
2. How does the dot product influence the work calculation?
3. How do we parametrize paths for more complex trajectories?
4. How does work change when force is not constant?
5. What other methods are there to compute work in physics problems?

**Tip**: The dot product between force and displacement vectors helps identify how much of the force is actually contributing to movement in the direction of motion.

A force F = 2i - yj, with x in meters, acts on a particle as it moves from coordinates (-1,5) to (2,-4). What is the work done?

Learn how to calculate the work done by a force F = 2i - yj on a particle as it moves from coordinates (-1, 5) to (2, -4). This problem covers vector calculus concepts like parametrization of paths and the dot product. Suitable for undergraduate-level physics and mathematics students.

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