Examples of fixed pulley?
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Actually Welcome to the concept of Efficiency.
Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%
The efficiency is => 22% => 22/100.
so we get as,
E = W(output) /W(input)
hence, W(output) = E x W(input)
so we get as,
W(output) = (22/100) x 2.2 x 10^7
=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7
hence, W(output) = 4.84 x 10^6 J
The useful work done on the mass is 4.84 x 10^6 J
(a)
Let's take the initial direction (before the collision) of the cue ball has positive x-direction.
Along the y-direction, the total initial momentum is zero:
Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:
where
m is the mass of each ball
is the velocity of the cue ball after the collision
is the velocity of the second ball after the collision
is the angle of the cue ball with the x-axis
is the angle of the second ball
Solving for , we find the angle between the direction of motion of the second ball and the original direction of motion:
(b) 6.69 m/s
To find the original speed of the cue ball, we analyze the situation along the horizontal direction.
First, we calculate the total momentum along the x-direction after the collision, which is:
The initial total momentum along the x-direction as
where
m is the mass of the cue ball
is the initial velocity of the cue ball
The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u: