vf = 10 m/s. A ball with mass of 4kg and a impulse given of 28N.s with a intial velocity of 3m/s would have a final velocity of 10 m/s.
The key to solve this problem is using the equation I = F.Δt = m.Δv, Δv = vf - vi.
The impulse given to the ball with mass 4Kg is 28 N.s. If the ball were already moving at 3 m/s, to calculate its final velocity:
I = m(vf - vi) -------> I = m.vf - m.vi ------> vf = (I + m.vi)/m ------> vf = I/m + vi
Where I 28 N.s, m = 4 Kg, and vi = 3 m/s
vf = (28N.s/4kg) + 3m/s = 7m/s + 3m/s
vf = 10 m/s.
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For the work-energy theorem, the work needed to stop the bus is equal to its variation of kinetic energy:

where
W is the work
Kf is the final kinetic energy of the bus
Ki is the initial kinetic energy of the bus
Since the bus comes at rest, its final kinetic energy is zero:

, so the work done by the brakes to stop the bus is

And the work done is negative, because the force applied by the brake is in the opposite direction to that of the bus motion.
Answer:
Calculating Coefficient of friction is 0.229.
Force is 4.5 N that keep the block moving at a constant speed.
Explanation:
We know that speed expression is as
.
Where,
is initial speed, V is final speed, ∆s displacement and a acceleration.
Given that,
=3 m/s, V = 0 m/s, and ∆s = 2 m
Substitute the values in the above formula,

0 = 9 - 4a
4a = 9

is the acceleration.
Calculating Coefficient of friction:


Compare the above equation

Cancel "m" common term in both L.H.S and R.H.S





Hence coefficient of friction is 0.229.
calculating force:


F = 4.5 N
Therefore, the force would be <u>4.5 N</u> to keep the block moving at a constant speed across the floor.
Answer: Load divided by it effort
Explanation:
Mechanical advantage of any machine is its load divided by its effort