An 800kg police boat slows down uniformly from 50km/h (E) to 20km/h(E) as it enters a harbour. If the boat slows down over a 30m distance, what is the force of friction of the boat? (Hint: you will need to convert the velocities into m/s) please show your work.
To solve this problem, we can use the equation:
force (F) = mass (m) * acceleration (a)
In this case, the boat slows down uniformly, so the acceleration is constant. To find the acceleration, we can use the equation:
acceleration (a) = (final velocity - initial velocity) / time
However, we don't have the time information given in the question. But we have the distance traveled (30m) and the initial and final velocities. We can use kinematic equations to find the time taken.
The kinematic equation we can use is:
final velocity squared = initial velocity squared + 2 * acceleration * distance
First, we need to convert the velocities from km/h to m/s.
Given:
Initial velocity (u) = 50 km/h (east) = 50 * (1000 / 3600) = 13.89 m/s (east)
Final velocity (v) = 20 km/h (east) = 20 * (1000 / 3600) = 5.56 m/s (east)
Distance (s) = 30 m
Now, let's calculate the acceleration using the kinematic equation:
v^2 = u^2 + 2 * a * s
(5.56)^2 = (13.89)^2 + 2 * a * 30
30.88 = 192.72 + 60 * a
Rearranging the equation:
60 * a = 30.88 - 192.72
60 * a = -161.84
a = -161.84 / 60
a ≈ -2.70 m/s^2
The negative sign indicates deceleration or slowing down.
Now, we have all the required information to find the force of friction:
force (F) = mass (m) * acceleration (a)
Given:
Mass (m) = 800 kg
Acceleration (a) = -2.70 m/s^2
F = 800 kg * (-2.70 m/s^2)
F ≈ -2160 N
The force of friction acting on the boat is approximately 2160 Newtons in the direction opposite to its motion.
One of our primary physics tutors is on vacation with his family for Thanksgiving week.
You'll need to be even more patient. Sorry.