Question

An 800 kg police boat slows down uniformly from 50 km/h [E] to 20 km/h [E] as it enters a harbor. If the boat slows down over a 30 m distance, what is the force of friction on the boat? (Hint: You will need to convert the velocities into m/s.)

Answer 1

Answer on Question #53487, Physics / Other

Task: An 800 kg police boat slows down uniformly from 50 km/h [E] to 20 km/h [E] as it enters a harbor. If the boat slows down over a 30 m distance, what is the force of friction on the boat? (Hint: You will need to convert the velocities into m/s.)

Answer:

The acceleration of the boat is found from the equation:

$$v^2 = v_0^2 + 2(a*s)$$

, where $v$ – the final velocity, $v_0$ – the initial velocity and $s$ – the distance.

After converting the velocities into m/s:

$$v_0 = 50\ \mathrm{km/h} = 50000 / 3600\ \mathrm{m/s} = 13.89\ \mathrm{m/s}$$

and

$$v = 20\ \mathrm{km/h} = 20000 / 3600\ \mathrm{m/s} = 5.56\ \mathrm{m/s}$$

the acceleration equals: $a = (v^2 - v_0^2)/(2s) = (30.86 - 192.93)/60\ \mathrm{m}\ \mathrm{s}^{-2} = -2.701\ \mathrm{m}\ \mathrm{s}^{-2}$.

The force of friction is defined:

$$F = m*a = 800\ \mathrm{kg} \times (-2.7\ \mathrm{m}\ \mathrm{s}^{-2}) = -2160.96\ \mathrm{N}$$

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