Work-Energy Theorem (Grade 12 NSC Matric Physical Sciences): Revision Notes

Worked Example: Falling Brick

QUESTION A 1 kg brick is dropped from a height of 10 m. Calculate the work done on the brick between the moment it is released and when it hits the ground. Assume air resistance can be neglected.

SOLUTION

Step 1: Determine what is given and what is required

• Mass of brick: $m = 1$ kg
• Initial height: $h_i = 10$ m
• Final height: $h_f = 0$ m
• Required: work done on the brick

Step 2: Determine the approach The brick falls freely, so energy is conserved. The work done equals the change in kinetic energy. Initially, the brick has no kinetic energy because it starts from rest. When it hits the ground, all potential energy converts to kinetic energy.

Step 3: Determine the brick's potential energy at $h_i$ $E_p = mgh = (1)(9.8)(10) = 98$ J

Step 4: Determine the work done Initially: $E_{k,i} = 0$ J and $E_{p,i} = 98$ J Finally: $E_{k,f} = 98$ J and $E_{p,f} = 0$ J

From the work-energy theorem: $W_{net} = \Delta E_k = E_{k,f} - E_{k,i} = 98 - 0 = 98$ J

Therefore, 98 J of work was done on the brick.

Worked Example: Car Braking

QUESTION The driver of a 1000 kg car travelling at 16.7 m·s⁻¹ applies the brakes when seeing a red light. The brakes provide a frictional force of 8000 N. Determine the stopping distance of the car.

SOLUTION

Step 1: Determine what is given and what is required

• Mass of car: $m = 1000$ kg
• Initial speed: $v = 16.7$ m·s⁻¹
• Frictional force of brakes: $F = 8000$ N
• Required: stopping distance

Step 2: Determine the approach We apply the work-energy theorem. All the car's kinetic energy is lost to friction, so the change in kinetic energy equals the work done by the brakes.

Step 3: Determine the kinetic energy of the car $E_k = \frac{1}{2}mv^2 = \frac{1}{2}(1000)(16.7)^2 = 139\,445$ J

Step 4: Determine the work done The stopping distance is $\Delta x_0$. Since the applied force and displacement are in opposite directions, $\theta = 180°$.

$W = F\Delta x \cos \theta = (8000)(\Delta x_0) \cos(180°) = (8000)(\Delta x_0)(-1) = (-8000)(\Delta x_0)$

Step 5: Apply the work-energy theorem The change in kinetic energy equals the work done: $\Delta E_k = W_{net}$ $E_{k,f} - E_{k,i} = (-8000)(\Delta x_0)$ $0 - 139\,445 = (-8000)(\Delta x_0)$ $\therefore \Delta x_0 = \frac{139445}{8000} = 17.4$ m

Step 6: Write the final answer The car stops in 17.4 m.

Worked Example: Block on an Inclined Plane

QUESTION A 2 kg block is pulled up along a smooth incline of length 10 m and height 5 m by applying a non-conservative force. At the end of the incline, the block is released from rest to slide down to the bottom. Find:

1. the work done by the non-conservative force
2. the kinetic energy of the block at the end of the round trip
3. the speed at the end of the round trip

SOLUTION

Step 1: Analyse the given information Three forces act on the block while going up:

1. Weight of the block: $F_g = mg$
2. Normal force: $N$ (applied by the block)
3. Non-conservative force: $F$

During the downward journey, only two forces act: weight and normal force.

Step 2: Work done by non-conservative force during round trip The work done by the non-conservative force is $W_F$.

Since $F$ acts only during the upward journey and the block is released at the end of the upward journey: $W_F = W_{F(up)} + W_{F(down)} = W_{F(up)} + 0 = W_{F(up)}$

The kinetic energies at the beginning and end of the upward motion are both zero.

Since the net work done during upward motion is zero: $W_{net} = W_{F(up)} + W_{g(up)} + W_{N(up)}$ $0 = W_{F(up)} + W_{g(up)} + W_{N(up)}$

Step 3: Work done by normal force during upward motion The normal force is perpendicular to the direction of motion, so it does no work: $W_{F(up)} = -W_{g(up)}$

Step 4: Work done by gravity during upward motion For upward motion, the component of gravitational force parallel to the slope $(F_{gx})$ opposes the motion. The angle between force and displacement is $\theta = 180°$, so $\cos \theta = -1$.

The magnitude of $F_{gx} = mg \sin \alpha$. The work done by gravity during upward motion is: $W_{g(up)} = F_{gx} \Delta x \cos \theta = mg \sin \alpha \cdot \Delta x \cos \theta = (2)(9.8)\left(\frac{5}{10}\right)(10) \cos(180°) = -98$ J

Therefore, the work done by the non-conservative force during the round trip is: $W_F = W_{F(up)} = -W_{g(up)} = -(-98) = 98$ J

Step 5: Kinetic energy at the end of round trip Using the work-energy theorem for the entire motion: $W_{(round trip)} = E_{k,f} = E_{k,f} - E_{k,i} = E_{k,f} - 0 = E_{k,f}$

The total work done during the round trip includes work by all forces: $W_{(round trip)} = W_F + W_g + W_N = W_F + W_{g(up)} + W_{N(up)} = (98) + (0) + (0) = 98$ J

Therefore: $E_{k,f} = 98$ J

Step 6: Speed of the block $E_{k,f} = \frac{1}{2}mv^2$ $\frac{1}{2}mv^2 = E_{k,f}$

$v = \pm\sqrt{\frac{2E_{k,f}}{m}} = \pm\sqrt{\frac{2(98)}{2}} = \pm\sqrt{98} = \pm 9.90$ m·s⁻¹

The speed at the end of the round trip is 9.90 m·s⁻¹.

The positive work done by the non-conservative force is cancelled by the negative work done by gravity during downward motion. The net work results in a change in kinetic energy as per the work-energy theorem.