Mathematics 9709/41 — May/June 2010

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

questions

marks

minutes

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

Q14MEnergy, Work and PowerNewton's Laws of MotionFree sample

A car of mass $1150\text{ kg}$ travels up a straight hill inclined at $1.2^\circ$ to the horizontal. The resistance to motion of the car is $975\text{ N}$ . Find the acceleration of the car at an instant when it is moving with speed $16\text{ m s}^{-1}$ and the engine is working at a power of $35\text{ kW}$ .

DifficultyMedium

Worked solution

Approach

The driving force supplied by the engine is found from $\text{power} = \text{force} \times \text{speed}$ . Then resolve the forces parallel to the slope and apply Newton's second law to find the acceleration.

Working

Let $DF$ be the driving force. Using $P = DF\,v$ :

DF = \frac{P}{v} = \frac{35000}{16} = 2187.5\ \text{N}

Resolving parallel to the slope, taking the direction of motion (up the slope) as positive:

driving force $DF$ up the slope,
component of weight down the slope $= mg\sin 1.2^\circ$ ,
resistance $975\ \text{N}$ down the slope.

Using $g = 10\ \text{m s}^{-2}$ , apply Newton's second law:

DF - 1150g\sin 1.2^\circ - 975 = 1150a

2187.5 - 1150(10)\sin 1.2^\circ - 975 = 1150a

Since $\sin 1.2^\circ = 0.02094$ , this becomes

2187.5 - 240.8 - 975 = 1150a

971.7 = 1150a

a = \frac{971.7}{1150} = 0.845\ \text{m s}^{-2}

Answer

The acceleration of the car is $0.845\ \text{m s}^{-2}$ .

Final answer

0.845 m s^-2

Detailed explanation

Walkthrough

The engine power tells us how quickly the engine is doing work. At the instant when the car has speed $16\ \text{m s}^{-1}$ , the driving force is obtained from $P = Fv$ , so

F = \frac{P}{v} = \frac{35000}{16} = 2187.5\ \text{N}

Remember to convert $35\ \text{kW}$ into $35000\ \text{W}$ before using this formula.

Next, identify all forces acting along the slope. The driving force acts up the slope. The weight $mg$ acts vertically downwards, so its component along the slope is $mg\sin 1.2^\circ$ , directed down the slope. The resistance of $975\ \text{N}$ also acts down the slope.

Taking the up-slope direction as positive, the resultant force along the slope is

DF - 1150g\sin 1.2^\circ - 975

By Newton's second law, this resultant force equals mass times acceleration:

DF - 1150g\sin 1.2^\circ - 975 = 1150a

Substituting the driving force and $g = 10\ \text{m s}^{-2}$ gives the acceleration.

Key Takeaways

The relation $P = Fv$ connects engine power to the driving force at a given speed.
On an inclined slope, the component of weight along the slope is $mg\sin\theta$ .
Newton's second law is applied by taking the resultant force in the direction of motion.
Always use consistent units: watts, newtons, metres per second and kilograms.

Common Mistakes

Using $35$ instead of $35000$ in the power formula.
Forgetting to include the component of the car's weight down the slope.
Using $mg\cos\theta$ instead of $mg\sin\theta$ for the component along the slope.
Mixing up signs: the driving force is up the slope, while the weight component and resistance are down the slope.
Quoting the acceleration without showing the Newton's second law equation, as the mark scheme requires the method to be shown.

Things to Be Careful About

The power formula $P = Fv$ gives the force only when the force and velocity are in the same direction, which is true here.
Take care to convert kilowatts to watts before calculating.
The normal reaction does not appear in this calculation because it is perpendicular to the motion.
Use the value of $g$ stated in the question; here the mark scheme is consistent with $g = 10\ \text{m s}^{-2}$ .
Check that all forces are resolved correctly parallel to the slope, not horizontally or vertically.

Techniques used

convert engine power and speed into driving force using P = Fvresolve the weight component down the slopeapply Newton's second law along the direction of motionsolve the resulting equation for the acceleration

The rest of this paper

6 more questions

Q2Kinematics of Motion in a Straight Line5M
Q3Forces and Equilibrium5M
Q4Forces and Equilibrium7M
Q5Energy, Work and Power7M
Q6Newton's Laws of Motion · Forces and Equilibrium11M
Q7Kinematics of Motion in a Straight Line11M

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