Question

Victoria draws back her bow 0.85m, ready to firean arrow. The spring constant of the bow elastic is800N/m. She draws the bow back in 0.2sFind the power that Victoria pulls the bow backwith.​

Answer 1

The power that Victoria pulls the bow back with is 1445 W.

Step-by-step explanation:

To calculate the power that Victoria pulls the bow back with, we can use the formula: P = W/t, where P is the power, W is the work done, and t is the time taken. In this case, the work done is equal to the potential energy stored in the bow. This can be calculated using the formula: PE = (1/2)kx^2, where PE is the potential energy, k is the spring constant, and x is the displacement.

First, let's calculate the potential energy stored in the bow: PE = (1/2)(800 N/m)(0.85 m)^2 = 289 J.

Next, we can calculate the power: P = 289 J / 0.2 s = 1445 W.

Answer 2

Approximately
`1.4 * 10^(3)\; {\rm W}` on average.

Step-by-step explanation:

Power is the rate at which work is being done. Divide the total amount of work that is done by duration to find the average power.

In this question, the average power associated with pulling the bow back can be found in the following steps:

• Find the elastic potential energy stored in the bow when the bow is fully pulled back.
• Deduce that the amount of work that the person did when pulling the bow back is equal to the change in the elastic potential energy stored in the bow.
• Divide the amount of work done by duration to find the average rate at which work is being done, which is equal to average power.

The elastic potential energy (
`\text{EPE}`) stored in an ideal spring is:

`\displaystyle (\text{EPE}) = (1)/(2)\, k\, x^(2)`, where:

• 
  `k` is the spring constant, and
• 
  `x` is the displacement of the spring from the equilibrium position.

For the bow in this question,
`k = 800\; {\rm N\cdot m^(-1)}` and
`x = 0.85\; {\rm m}` when the bow is fully pulled back.

Assuming that the work that has been done when pulling back the bow is entirely converted into the elastic potential energy of the bow, such that no energy is lost (e.g., as heat.) The amount of work that was done when pulling back the bow would be equal to the gain in the elastic potential energy,
`(1/2)\, k\, x^(2)`.

Divide the work that is done by the time required to achieve the work to find the average power,
`P`:

`\begin{aligned}P &= \frac{(\text{work done})}{t}\\ &= \frac{(\text{EPE})}{t} \\ &= ((1/2)\, k\, x^(2))/(t) \\ &= ((1/2)\, (800)\, (0.85)^(2))/((0.2))\; {\rm W} \\ &\approx 1.4 * 10^(3)\; {\rm W}\end{aligned}`.