Physics 5054/21 — October/November 2016

Walkthrough

A resultant force causes an acceleration. Newton's second law tells us that

$F = ma$

where $F$ is the resultant force in newtons, $m$ is the mass in kilograms and $a$ is the acceleration in $\text{m/s}^2$.

The car has mass $800\text{ kg}$ and acceleration $1.5\text{ m/s}^2$. Substituting:

$F = 800 \times 1.5 = 1200\text{ N}$

The answer is $1200\text{ N}$.

Key Takeaways

• Resultant force is the single force that has the same effect as all the actual forces added together.
• For a constant mass, $F = ma$.
• The unit of force is the newton.

Common Mistakes

• Forgetting the direction: the resultant force acts in the direction of the acceleration, forwards here.
• Using the initial speed of $25\text{ m/s}$; it is not needed in $F = ma$.
• Forgetting the unit N.

Things to Be Careful About

• The mass is already in kilograms and the acceleration in $\text{m/s}^2$, so no unit conversion is needed.
• 800 kg is the mass, not the weight. Weight would use $W = mg$ and is not relevant to horizontal acceleration.
• Show the equation and substitution because the first mark is a working mark (C1).

Walkthrough

The value calculated in (b)(i) is the resultant force, not the force produced by the engine alone. As the car moves, friction between the tyres and road, and air resistance/drag, act backwards on the car.

If we choose forward as positive, the resultant force is

$F_{\text{resultant}} = F_{\text{engine}} - F_{\text{friction/drag}}$

So for the resultant force to be $1200\text{ N}$ forward, the engine force must be $1200\text{ N}$ plus the friction/drag force. That is why the engine force is greater than the value in (b)(i).

The mark scheme credits two separate ideas: that friction/air resistance acts on the car, and that it opposes the force due to the engine.

Key Takeaways

• The resultant force is not usually the same as the applied force. It is the vector sum of all the forces.
• When a car accelerates, the engine must provide the resultant force AND overcome friction and air resistance.
• Resistive forces act opposite to the direction of motion.

Common Mistakes

• Saying only friction acts without saying it opposes motion or the engine force.
• Using the word gravity; gravity acts vertically and does not explain the horizontal engine force being larger.
• Confusing the engine force with the resultant force; they are different unless there is no friction or drag.

Things to Be Careful About

• This is an explain style question; both marking points need to be present.
• The answer should make the opposition clear: friction and air resistance oppose the forward force.
• Do not introduce terminal velocity here; the car is accelerating, not travelling at a constant terminal speed.

Walkthrough

The initial speed is $25\text{ m/s}$. During the $4.0\text{ s}$ the car accelerates at a constant $1.5\text{ m/s}^2$, so its speed increases by

$\Delta v = at = 1.5 \times 4.0 = 6.0\text{ m/s}$

The new speed is therefore

$v = 25 + 6.0 = 31\text{ m/s}$

This is the same as using the equation

$v = u + at$

with $u = 25\text{ m/s}$, $a = 1.5\text{ m/s}^2$ and $t = 4.0\text{ s}$.

Key Takeaways

• For uniform acceleration, change in speed is $at$.
• Final speed = initial speed + change in speed.
• The units must match: using m/s, m/s^2 and seconds gives m/s.

Common Mistakes

• Forgetting to add the initial speed $25\text{ m/s}$ and writing only $6.0\text{ m/s}$.
• Subtracting the change instead of adding it.
• Quoting the final answer without the unit.
• Using the mass of the car; it is not needed in this kinematics calculation.

Things to Be Careful About

• The acceleration is constant, so the simple equation is valid.
• 4.0 s and 1.5 m/s^2 are in compatible SI units; no conversion is required.
• If the question asked for velocity rather than speed, a direction would also be needed. Here it asks for speed, so a magnitude and unit are enough.