(a) The speed of a moving bus reduced from 45m/s to 5m/s with a uniform retardation of 10m/s\(^2\). Calculate the distance covered.
(b) A bucket full of water with mass 16kg is pulled out of a well with a light inextensible rope. Find its acceleration when the tension in the rope is 240N. [Take g= 10m/s\(^2\)]
(a) The initial velocity of the bus is 45 m/s and the final velocity is 5 m/s, and the retardation is 10 m/s2. We can use the equation of motion to find the distance covered. The equation is:
v2 = u2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance covered.
Substituting the given values, we get:
52 = 452 + 2(-10)s
25 = 2025 - 20s
20s = 2000
s = 100 meters
Therefore, the distance covered by the bus is 100 meters.
(b) The weight of the bucket is given by W = mg, where m is the mass of the bucket and g is the acceleration due to gravity. The tension in the rope is equal to the weight of the bucket, so we have:
T = W = mg
The net force acting on the bucket is given by F = T - mg, where F is the force accelerating the bucket.
Using Newton's second law of motion, F = ma, where a is the acceleration of the bucket. Substituting the values, we get:
ma = T - mg
a = (T - mg) / m
Substituting the given values, we get:
a = (240 - 16 x 10) / 16
a = 10 m/s2
Therefore, the acceleration of the bucket is 10 m/s2.