A force of 5.0 N acts on a 15 kg body initially at rest. Compute the work done by the force in (a) the first, (b) the second, and (c) the third seconds and (d) the instantaneous power due to the force at the end of the third second.

Answer

To solve the problem step by step, we will break it down into parts (a), (b), (c), and (d) as requested.

### Given:
- Force (F) = 5.0 N
- Mass (m) = 15 kg
- Initial velocity (u) = 0 m/s (body is initially at rest)

### Step 1: Calculate the acceleration.
Using Newton's second law, we can find the acceleration (a) using the formula:
\[ a = \frac{F}{m} \]

Substituting the values:
\[ a = \frac{5.0 \, \text{N}}{15 \, \text{kg}} = \frac{1}{3} \, \text{m/s}^2 \]

### Step 2: Calculate the work done in the first second (part a).
The displacement (s) in the first second can be calculated using the formula:
\[ s = ut + \frac{1}{2} a t^2 \]
Since the initial velocity (u) is 0:
\[ s = 0 + \frac{1}{2} \left(\frac{1}{3}\right)(1^2) = \frac{1}{6} \, \text{m} \]

Now, work done (W) is given by:
\[ W = F \cdot s \]
Substituting the values:
\[ W = 5.0 \, \text{N} \cdot \frac{1}{6} \, \text{m} = \frac{5}{6} \, \text{J} \approx 0.83 \, \text{J} \]

### Step 3: Calculate the work done in the second second (part b).
For the second second, we can use the formula for displacement during the nth second:
\[ s_n = u + \frac{a}{2}(2n - 1) \]
Substituting \( n = 2 \):
\[ s_2 = 0 + \frac{1}{2} \left(\frac{1}{3}\right)(2 \cdot 2 - 1) = \frac{1}{2} \left(\frac{1}{3}\right)(3) = \frac{1}{2} \, \text{m} \]

Now, calculate the work done:
\[ W = F \cdot s_2 = 5.0 \, \text{N} \cdot \frac{1}{2} \, \text{m} = 2.5 \, \text{J} \]

### Step 4: Calculate the work done in the third second (part c).
For the third second, using the same formula:
\[ s_3 = u + \frac{a}{2}(2n - 1) \]
Substituting \( n = 3 \):
\[ s_3 = 0 + \frac{1}{2} \left(\frac{1}{3}\right)(2 \cdot 3 - 1) = \frac{1}{2} \left(\frac{1}{3}\right)(5) = \frac{5}{6} \, \text{m} \]

Now, calculate the work done:
\[ W = F \cdot s_3 = 5.0 \, \text{N} \cdot \frac{5}{6} \, \text{m} = \frac{25}{6} \, \text{J} \approx 4.17 \, \text{J} \]

### Step 5: Calculate the instantaneous power at the end of the third second (part d).
First, we need to find the final velocity (v) at the end of the third second using:
\[ v = u + at \]
Substituting \( t = 3 \):
\[ v = 0 + \left(\frac{1}{3}\right)(3) = 1 \, \text{m/s} \]

Now, the instantaneous power (P) is given by:
\[ P = F \cdot v \]
Substituting the values:
\[ P = 5.0 \, \text{N} \cdot 1 \, \text{m/s} = 5 \, \text{W} \]

### Final Answers:
(a) Work done in the first second: \( \approx 0.83 \, \text{J} \)
(b) Work done in the second second: \( 2.5 \, \text{J} \)
(c) Work done in the third second: \( \approx 4.17 \, \text{J} \)
(d) Instantaneous power at the end of the third second: \( 5 \, \text{W} \)