9. An 800 kg police boat slows down uniformly from 50 km/h [E] to 20 km/h (E) as it enters a

harbour. 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.)

2,158N

Explanation:

According to the second newtons law of motion, Force (F) = mass (m) * acceleration (a) ... 1

Given mass to be 800kg.

We need to get the acceleration of the body using one of the equations of motion.

v² = u²+2as ... (2) where;

v = final velocity = 50km/hr

u = initial velocity = 20km/hr

a is the acceleration

s is the distance covered by the boat = 30m

Converting 50km/hr and 20km/hr to m/s, this gives us

50km*1000m/1hr*3600s

= 50000/3600

= 13.9m/s

Similarly,

20km/hr = 20000m/3600s

= 5.6m/s

Substituting this values into the equation 2 to get acceleration, we have

13.9² = 5.6²+2a (30)

193.21 = 31.36+60a

60a = 193.21-31.36

60a = 161.85

a = 2.69m/s²

To get the force on the both we substitute the value of mass and acceleration in equation 1

F = ma

F = 800*2.69

F = 2,158N

Therefore the force of friction on the boat is 2,158N