Question1 ■ Mark for Review
A hawk with mass 1 kg dives straight downward. At time $t_{1}$ , the hawk has a speed of $20 \mathrm{~m} / \mathrm{s}$ . At time $t_{2}$ , the hawk has a speed of $50 \mathrm{~m} / \mathrm{s}$ . The change in the kinetic energy of the hawk between $t_{1}$ and $t_{2}$ is most nearly (A) 200 J (B) 450 J (C) 1050 J (D) 1250 J

Studdy Solution

STEP 1

1. The mass of the hawk is 1 kg.
2. The initial speed of the hawk at time $t_1$ is $20 \, \mathrm{m/s}$ .
3. The final speed of the hawk at time $t_2$ is $50 \, \mathrm{m/s}$ .
4. We are asked to find the change in kinetic energy of the hawk between $t_1$ and $t_2$ .

STEP 2

1. Recall the formula for kinetic energy.
2. Calculate the initial kinetic energy at $t_1$ .
3. Calculate the final kinetic energy at $t_2$ .
4. Determine the change in kinetic energy.

STEP 3

Recall the formula for kinetic energy, which is given by:
$KE = \frac{1}{2} m v^2$
where $m$ is the mass and $v$ is the velocity.

STEP 4

Calculate the initial kinetic energy at $t_1$ using the formula:
$KE_1 = \frac{1}{2} \times 1 \, \mathrm{kg} \times (20 \, \mathrm{m/s})^2$
$KE_1 = \frac{1}{2} \times 1 \times 400$
$KE_1 = 200 \, \mathrm{J}$

STEP 5

Calculate the final kinetic energy at $t_2$ using the formula:
$KE_2 = \frac{1}{2} \times 1 \, \mathrm{kg} \times (50 \, \mathrm{m/s})^2$
$KE_2 = \frac{1}{2} \times 1 \times 2500$
$KE_2 = 1250 \, \mathrm{J}$

STEP 6

Determine the change in kinetic energy by subtracting the initial kinetic energy from the final kinetic energy:
$\Delta KE = KE_2 - KE_1$
$\Delta KE = 1250 \, \mathrm{J} - 200 \, \mathrm{J}$
$\Delta KE = 1050 \, \mathrm{J}$
The change in kinetic energy of the hawk between $t_1$ and $t_2$ is most nearly:
$\boxed{1050 \, \mathrm{J}}$

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