. A 3.0 kg mass of body initially at rest which enis move forward in horizontal direction ofSanforce 0.60 N on a smooth table. Find the work done 9.0 J the force in 9.05 s and show that this equals the change in the kinetic energy of the body.

We are tasked with finding the work done, force, and showing that the work done is equal to the change in kinetic energy for a 3.0 kg mass, given that it starts from rest and moves under the influence of a 0.60 N force on a smooth table over a time period of 9.05 seconds. 

Let's break down the problem step by step.

### Step 1: Work Done (W)
Work done by a force is given by the equation:

\[
W = F \cdot d
\]

where:
- \( F \) is the applied force (0.60 N),
- \( d \) is the displacement of the object.

We do not yet have the displacement \( d \), but we can find it using the equations of motion, since the table is smooth (no friction) and the body starts from rest.

### Step 2: Finding the Acceleration (a)
Using Newton's second law, the acceleration can be calculated as:

\[
a = \frac{F}{m}
\]

where:
- \( F = 0.60 \, \text{N} \),
- \( m = 3.0 \, \text{kg} \).

\[
a = \frac{0.60 \, \text{N}}{3.0 \, \text{kg}} = 0.20 \, \text{m/s}^2
\]

### Step 3: Displacement (d)
The displacement of the body can be found using the kinematic equation:

\[
d = v_i t + \frac{1}{2} a t^2
\]

Since the body starts from rest, \( v_i = 0 \), so the equation simplifies to:

\[
d = \frac{1}{2} a t^2
\]

Substitute \( a = 0.20 \, \text{m/s}^2 \) and \( t = 9.05 \, \text{seconds} \):

\[
d = \frac{1}{2} \times 0.20 \, \text{m/s}^2 \times (9.05 \, \text{s})^2
\]

Let's calculate this value:

\[
d = \frac{1}{2} \times 0.20 \times 81.9025 = 8.19 \, \text{m}
\]

### Step 4: Work Done (W)
Now that we have the displacement \( d = 8.19 \, \text{m} \), we can find the work done:

\[
W = F \cdot d = 0.60 \, \text{N} \times 8.19 \, \text{m} = 4.91 \, \text{J}
\]

### Step 5: Kinetic Energy (KE)
The kinetic energy of the body can be calculated using the equation:

\[
KE = \frac{1}{2} m v^2
\]

To find the final velocity \( v \), we use the equation of motion:

\[
v = v_i + at
\]

Since \( v_i = 0 \), we have:

\[
v = at = 0.20 \, \text{m/s}^2 \times 9.05 \, \text{s} = 1.81 \, \text{m/s}
\]

Now calculate the kinetic energy:

\[
KE = \frac{1}{2} \times 3.0 \, \text{kg} \times (1.81 \, \text{m/s})^2
\]

\[
KE = \frac{1}{2} \times 3.0 \times 3.2761 = 4.91 \, \text{J}
\]

### Conclusion:
The work done by the force (4.91 J) is equal to the change in kinetic energy of the body (4.91 J). This confirms the work-energy theorem.

---

Would you like any further details or have any questions?

Here are 5 related questions for further exploration:
1. How would the result change if there was friction on the table?
2. What is the significance of the work-energy theorem in physics?
3. How does increasing the mass affect the final kinetic energy for the same force?
4. What if the initial velocity wasn't zero? How would that impact the work done?
5. Can you explain why the smooth table (no friction) was important in this problem?

**Tip**: When solving motion problems, always check if the initial velocity is zero—it simplifies the equations a lot!

A 3.0 kg mass of body initially at rest is moved forward in the horizontal direction by a force of 0.60 N on a smooth table. Find the work done (9.0 J), the force in 9.05 s, and show that this equals the change in the kinetic energy of the body.

Calculate the work done by a 0.60 N force moving a 3.0 kg mass on a smooth table. Explore the work-energy theorem and how the work done (4.91 J) matches the change in kinetic energy over 9.05 seconds. Ideal for physics students in Grades 9-12.

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