Q.A body of mass $2\ \text{kg}$ initially at rest moves under the action of an applied horizontal force of $7\ \text{N}$ on a table with coefficient of kinetic friction $= 0.1$. Compute the

Concept: Work-Energy Theorem — the net work done on a body equals its change in kinetic energy.

Step 1: Find acceleration and distance.

Friction force: $f_k = \mu_k mg = 0.1 \times 2 \times 9.8 = 1.96\ \text{N}$ (opposing motion).

Net force: $F_{\text{net}} = 7 - 1.96 = 5.04\ \text{N}$.

Acceleration: $a = F_{\text{net}}/m = 5.04/2 = 2.52\ \text{m/s}^2$.

Distance in $10\ \text{s}$ from rest: $s = \frac{1}{2} a t^2 = \frac{1}{2} \times 2.52 \times 100 = 126\ \text{m}$.

Step 2: Compute works.

1. Work by applied force: $W_{\text{app}} = F_{\text{app}} \cdot s = 7 \times 126 = 882\ \text{J}$.
2. Work by friction: $W_{\text{fric}} = -f_k \cdot s = -1.96 \times 126 = -246.96\ \text{J}$.
3. Work by net force: $W_{\text{net}} = F_{\text{net}} \cdot s = 5.04 \times 126 = 635.04\ \text{J}$. Step 3: Change in kinetic energy. Final velocity: $v = a t = 2.52 \times 10 = 25.2\ \text{m/s}$. $\Delta K = \frac{1}{2} m v^2 - 0 = \frac{1}{2} \times 2 \times (25.2)^2 = 635.04\ \text{J}$. Interpretation: $W_{\text{net}} = \Delta K$ confirms the Work-Energy Theorem. The applied force does positive work, friction does negative work, and their sum equals the gain in kinetic energy.
   ✓Final answer

   1. $882\ \text{J}$.
   2. $-246.96\ \text{J}$.
   3. $635.04\ \text{J}$.
   4. $635.04\ \text{J}$ — net work equals change in kinetic energy.

The key to this problem is the Work-Energy Theorem: the net work done on a body equals its change in kinetic energy. But to compute individual works, we first need the displacement — which requires finding the acceleration from the net force.

Let's break it down.

1. Find the net force and acceleration

The applied force is $F_{\text{app}} = 7\ \text{N}$ forward.

Kinetic friction opposes motion: $f_k = \mu_k N$, where $N = mg$ (since the surface is horizontal).

Mass $m = 2\ \text{kg}$, $g = 9.8\ \text{m/s}^2$, $\mu_k = 0.1$.

Net force:

Acceleration:

2. Displacement in 10 seconds

Body starts from rest ($u = 0$). Using $s = ut + \frac12 a t^2$:

3. Work done by each force

1. Work by applied force Force and displacement are in the same direction:

   $W_{\text{app}} = F_{\text{app}} \cdot s = 7 \times 126 = 882\ \text{J}$

2. Work by friction Friction opposes motion, so $\theta = 180^\circ$:

   $W_{\text{fric}} = f_k \cdot s \cdot \cos 180^\circ = 1.96 \times 126 \times (-1) = -246.96\ \text{J} \approx -247\ \text{J}$

3. Work by net force Either sum the works:

   $W_{\text{net}} = 882 + (-247) = 635\ \text{J}$

   Or directly: $F_{\text{net}} \times s = 5.04 \times 126 = 635.04\ \text{J} \approx 635\ \text{J}$

4. Change in kinetic energy

Final velocity after 10 s:

Initial KE = 0. Final KE:

So change in KE:

5. Interpretation

Notice: $W_{\text{net}} = 635\ \text{J}$ and $\Delta \text{KE} = 635\ \text{J}$ — exactly equal. This is the Work-Energy Theorem in action.

The applied force does 882 J of work, but 247 J is "lost" to friction (converted to heat), leaving 635 J to increase the body's kinetic energy.

Concept: Work-Energy Theorem Combined with Kinematics

Step 1: Find the net force and acceleration.

Step 2: Find the displacement in $10\,\text{s}$ from rest.

Step 3: Compute the work done by each force.

Step 4: Verify against the change in kinetic energy.

Final Answer:

(a) 882 J, (b) $-246.96$ J, (c) 635.04 J, (d) 635.04 J — confirming $W_{\text{net}}=\Delta K$