Question

A constant force of 10N acts on a 2.0kg object initially at rest on a horizontal surface and gives it 90J of kinetic energy. Calculate :
a. (I) The time taken
(II) The power

Answer 1

To solve the problem, we can use the fact that work done on an object is equal to the change in kinetic energy, and we can also relate power to work done over time.

### Given:
- Force \( F = 10 \, \text{N} \)
- Mass \( m = 2.0 \, \text{kg} \)
- Kinetic Energy \( KE = 90 \, \text{J} \)

### (a) (I) Time Taken

First, we can calculate the acceleration \( a \) using Newton's second law:

\[
F = m \cdot a \quad \Rightarrow \quad a = \frac{F}{m} = \frac{10 \, \text{N}}{2.0 \, \text{kg}} = 5 \, \text{m/s}^2
\]

Next, we can use the work-energy principle. The work done \( W \) by the force is equal to the change in kinetic energy (which is given as 90 J).

We can also calculate the distance \( s \) traveled using the work done:

\[
W = F \cdot s \quad \Rightarrow \quad 90 \, \text{J} = 10 \, \text{N} \cdot s \quad \Rightarrow \quad s = \frac{90 \, \text{J}}{10 \, \text{N}} = 9 \, \text{m}
\]

Now we can use the kinematic equation that relates acceleration, distance, initial velocity (which is 0), and time:

\[
s = ut + \frac{1}{2}at^2
\]

Since the initial velocity \( u = 0 \):

\[
9 = 0 \cdot t + \frac{1}{2}(5)t^2 \quad \Rightarrow \quad 9 = \frac{5}{2}t^2 \quad \Rightarrow \quad 9 = 2.5t^2 \quad \Rightarrow \quad t^2 = \frac{9}{2.5} \quad \Rightarrow \quad t^2 = 3.6 \quad \Rightarrow \quad t = \sqrt{3.6} \approx 1.897 \, \text{s}
\]

So, the time taken is approximately:

\[
t \approx 1.9 \, \text{s}
\]

### (a) (II) Power

Power \( P \) is defined as the work done over time:

\[
P = \frac{W}{t} = \frac{90 \, \text{J}}{1.897 \, \text{s}} \approx 47.4 \, \text{W}
\]

### Summary of Answers:
- (I) Time taken: \( t \approx 1.9 \, \text{s} \)
- (II) Power: \( P \approx 47.4 \, \text{W} \)