Dynamics: Question 10

Worked solution

Step 1: Find the acceleration

The car starts from rest, so $u=0$, and reaches $v=20\text{ m s}^{-1}$ in $t=8.0\text{ s}$:

$a = \frac{\Delta v}{t} = \frac{20-0}{8.0} = 2.5\text{ m s}^{-2}$

Step 2: Apply Newton’s second law

$F = ma = 800\times2.5 = 2000\text{ N}$

(Equivalently, using force as the rate of change of momentum: $\Delta p = m\Delta v = 800\times20=16000\text{ kg m s}^{-1}$, so $F=\Delta p/\Delta t = 16000/8.0=2000\text{ N}$. The same result.)

Why the other options are wrong

• A ($100\text{ N}$): comes from dividing the mass by the time ($800/8.0$), without ever using the car’s change in velocity.
• C ($7850\text{ N}$): this is the car’s weight, $mg=800\times9.81\approx7850\text{ N}$, a vertical force, unrelated to the horizontal resultant force needed to accelerate the car.
• D ($16000\text{ N}$): this is the change in momentum, $\Delta p=16000\text{ kg m s}^{-1}$, mistaken for the force itself. The working must still divide by the time interval to get a force.

Final answer

• The average resultant force is $\boxed{2000}\text{ N}$, option B.