Mathematics 9709/41 — May/June 2012

Cambridge AS Level · Mechanics · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Newton's Laws of Motion · Energy, Work and Power · Forces and Equilibrium · Kinematics of Motion in a Straight Line

Q14MNewton's Laws of MotionEnergy, Work and PowerFree sample

A car of mass $880\text{ kg}$ travels along a straight horizontal road with its engine working at a constant rate of $P\text{ W}$ . The resistance to motion is $700\text{ N}$ . At an instant when the car’s speed is $16\text{ m s}^{-1}$ its acceleration is $0.625\text{ m s}^{-2}$ . Find the value of $P$ .

DifficultyMedium

Worked solution

Approach

Let $F$ be the driving force produced by the engine. The resistance is $700\text{ N}$ , so the resultant horizontal force is $F - 700$ . Applying Newton's second law gives $F - 700 = 880 \times 0.625$ . Then use the power relation $P = Fv$ , where $v = 16\text{ m s}^{-1}$ .

Working

Newton's second law:

F - 700 = 880 \times 0.625

Evaluate the right-hand side:

880 \times 0.625 = 550

Hence:

F = 700 + 550 = 1250

The driving force is $1250\text{ N}$ .

The power delivered by the engine is:

P = Fv = 1250 \times 16

P = 20\,000\text{ W}

Answer

P = 20\,000\text{ W}

Final answer

P = 20000 W

Detailed explanation

Walkthrough

First identify the forces on the car. The engine pushes the car forwards with a driving force $F$ , and the resistance acts backwards with magnitude $700\text{ N}$ . Since the acceleration is forwards, the resultant force is $F - 700$ .

Apply Newton's second law $F_{\text{net}} = ma$ :

F - 700 = 880 \times 0.625

Since $880 \times 0.625 = 550$ , we get $F = 1250\text{ N}$ . This is the instantaneous driving force needed to produce that acceleration while overcoming the resistance.

Next, because the engine works at constant power, the instantaneous power is $P = Fv$ . It is the driving force, not the resultant force, that is multiplied by the speed. Substitute $F = 1250\text{ N}$ and $v = 16\text{ m s}^{-1}$ :

P = 1250 \times 16 = 20\,000\text{ W}

Key Takeaways

Newton's second law applies to the resultant (net) force, so resistance must be subtracted from the driving force.
For constant engine power, instantaneous power is $P = Fv$ , where $F$ is the driving force and $v$ is the current speed.
At constant power, the driving force changes as the car speeds up or slows down; here we only need the values at one instant.

Common Mistakes

Using the resultant force $F - 700$ in $P = Fv$ instead of the engine's driving force $F$ . Power is supplied by the engine, so only the engine's driving force appears in $P = Fv$ .
Forgetting to add the $700\text{ N}$ resistance when solving $F - 700 = 880 \times 0.625$ , giving the wrong driving force.
Arithmetic slip: $880 \times 0.625 = 550$ , not $55$ .
Quoting $P$ in newtons or leaving units inconsistent. Power is measured in watts (W), equivalent to J/s.

Things to Be Careful About

The acceleration and force directions are along the same horizontal line, so signs matter: take the direction of motion as positive, making resistance negative.
The value $0.625\text{ m s}^{-2}$ is the acceleration at the instant the speed is $16\text{ m s}^{-1}$ ; $P = Fv$ gives instantaneous power.
The mark scheme requires showing Newton's second law and the use of $P = (\text{DF})v$ for the method marks, so include both steps.

Techniques used

apply Newton's second law to horizontal motionfind the driving force from the net forceuse instantaneous power as force times speed

The rest of this paper

6 more questions

Q2Forces and Equilibrium5M
Q3Energy, Work and Power6M
Q4Kinematics of Motion in a Straight Line8M
Q5Energy, Work and Power · Forces and Equilibrium · Newton's Laws of Motion · Kinematics of Motion in a Straight Line8M
Q6Newton's Laws of Motion · Kinematics of Motion in a Straight Line9M
Q7Forces and Equilibrium10M

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