To solve the problem of finding the power of the gravitational force acting on a ball of mass 1 kg dropped from a tower at time \( t = 2 \) seconds, we can follow these steps:
### Step 1: Identify the given values
- Mass of the ball, \( m = 1 \, \text{kg} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
- Time, \( t = 2 \, \text{s} \)
### Step 2: Calculate the velocity of the ball at \( t = 2 \, \text{s} \)
Since the ball is dropped from rest, we can use the equation of motion:
\[
v = u + gt
\]
where \( u \) is the initial velocity (which is \( 0 \) since the ball is dropped).
Substituting the values:
\[
v = 0 + (10 \, \text{m/s}^2)(2 \, \text{s}) = 20 \, \text{m/s}
\]
### Step 3: Calculate the gravitational force acting on the ball
The gravitational force \( F \) can be calculated using the formula:
\[
F = mg
\]
Substituting the values:
\[
F = (1 \, \text{kg})(10 \, \text{m/s}^2) = 10 \, \text{N}
\]
### Step 4: Calculate the power of the gravitational force
The power \( P \) can be calculated using the formula:
\[
P = F \cdot v \cdot \cos(\theta)
\]
Since the force and the velocity are in the same direction (downward), the angle \( \theta = 0^\circ \) and \( \cos(0^\circ) = 1 \).
Thus, the power becomes:
\[
P = F \cdot v \cdot 1 = F \cdot v
\]
Substituting the values:
\[
P = (10 \, \text{N})(20 \, \text{m/s}) = 200 \, \text{W}
\]
### Final Answer
The power of the gravitational force at \( t = 2 \, \text{s} \) is \( 200 \, \text{W} \).
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