Mathematics 9709/41 — October/November 2014

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

questions

marks

minutes

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

Q14MEnergy, Work and PowerNewton's Laws of MotionFree sample

A car of mass $800\text{ kg}$ is moving on a straight horizontal road with its engine working at a rate of $22.5\text{ kW}$ . Find the resistance to the car's motion at an instant when the car's speed is $18\text{ m s}^{-1}$ and its acceleration is $1.2\text{ m s}^{-2}$ .

DifficultyMedium-Easy

Worked solution

Approach

Use the power–velocity relation $P = Fv$ to find the driving force $DF$ , then apply Newton's second law along the horizontal direction. The engine force acts forwards and the resistance acts backwards, so $DF - R = ma$ .

Working

Engine power:

P = 22.5\text{ kW} = 22500\text{ W}

Driving force from $P = DF \times v$ :

DF = \frac{P}{v} = \frac{22500}{18} = 1250\text{ N}

Newton's second law horizontally:

DF - R = ma

1250 - R = 800 \times 1.2 = 960

Solve for $R$ :

R = 1250 - 960 = 290\text{ N}

Answer

The resistance to the car's motion is $290\text{ N}$ .

Final answer

290 N

Detailed explanation

Walkthrough

The car is moving on a horizontal road, so we only need to consider horizontal forces: the driving force $DF$ from the engine and the resistance $R$ opposing the motion. The engine power is $22.5\text{ kW}$ , which is $22500\text{ W}$ because $1\text{ kW} = 1000\text{ W}$ .

For a constant power output, $P = Fv$ , so the driving force at this instant is

DF = \frac{P}{v} = \frac{22500}{18} = 1250\text{ N}.

Now apply Newton's second law in the direction of motion. The resultant horizontal force is $DF - R$ , since $R$ opposes the motion, and this must equal $ma$ :

DF - R = 800 \times 1.2 = 960.

Substitute $DF = 1250$ :

1250 - R = 960,

R = 1250 - 960 = 290\text{ N}.

The resistance is therefore $290\text{ N}$ .

Key Takeaways

Power and force are related by $P = Fv$ at the instant considered.
In horizontal motion with forward and backward forces, Newton's second law requires the resultant force: $F_{\text{resultant}} = ma$ .
Resistance is a force that acts opposite to the direction of motion, so it is subtracted from the driving force.
Units must be consistent: kilowatts must be converted to watts before using $P = Fv$ .

Common Mistakes

Forgetting to convert $22.5\text{ kW}$ to $22500\text{ W}$ , which would give a wrong driving force.
Writing Newton's second law as $DF + R = ma$ instead of $DF - R = ma$ . Since $R$ opposes the motion, it must be subtracted.
Quoting the resistance without showing the Newton's second law equation; the mark scheme awards a method mark for using three terms, so working must be shown.
Using $DF = P \times v$ instead of $DF = P/v$ .

Things to Be Careful About

The driving force is instantaneous because the power and speed are given at that instant.
Check the direction: the engine force is in the direction of motion, while the resistance opposes it.
The final answer must be in newtons.
If the acceleration were negative, the sign of $ma$ would change, but here acceleration is positive in the direction of motion.

Techniques used

convert engine power to driving force using P = Fvapply Newton's second law to horizontal motionsolve for the resistance force

The rest of this paper

6 more questions

Q2Newton's Laws of Motion · Forces and Equilibrium4M
Q3Forces and Equilibrium6M
Q4Kinematics of Motion in a Straight Line7M
Q5Forces and Equilibrium · Newton's Laws of Motion9M
Q6Newton's Laws of Motion · Kinematics of Motion in a Straight Line9M
Q7Energy, Work and Power11M

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