To solve this problem, we can use some fundamental concepts in physics.
Given Data:
- Force (F) = 10 N
- Mass (m) = 2.0 kg
- Kinetic Energy (KE) = 90 J
a. Time taken (t)
We can find the acceleration \(a\) using Newton's second law:
\[ F = ma \quad \Rightarrow \quad a = \frac{F}{m} = \frac{10 \text{ N}}{2.0 \text{ kg}} = 5 \text{ m/s}^2 \]
Next, we can use the work-energy principle to find the time taken. The work done by the constant force is equal to the change in kinetic energy. Since the object was initially at rest, the work done is equal to the kinetic energy gained:
\[ W = F \cdot d = KE \]
First, we can find the distance \(d\) traveled using the kinetic energy:
\[ 90 J = 10 N \cdot d \quad \Rightarrow \quad d = \frac{90 J}{10 N} = 9 \text{ m} \]
Now, we can find the time taken using the kinematic equation:
\[ d = ut + \frac{1}{2}at^2 \]
Given that the initial velocity \(u = 0\):
\[ d = \frac{1}{2}at^2 \quad \Rightarrow \quad 9 = \frac{1}{2} (5)t^2 \]
\[ 9 = \frac{5}{2}t^2 \quad \Rightarrow \quad t^2 = \frac{9 \times 2}{5} = \frac{18}{5} \quad \Rightarrow \quad t^2 = 3.6 \]
\[ t = \sqrt{3.6} \approx 1.897 \text{ seconds} \]
b. Power (P)
Power is defined as the work done over a time interval. The average power can be calculated by:
\[ P = \frac{W}{t} \]
Where \(W = 90 J\) (the work done):
\[ P = \frac{90 J}{1.897 s} \approx 47.4 \text{ W} \]
Summary of Results
- a. The time taken \(t \approx 1.897\) seconds
- b. The power \(P \approx 47.4\) Watts