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A machine which is 75% efficient uses 12 J of energy in lifting up a 1 kg mass through a certain distance. The mass is allowed to call through that distance. Find the velocity of that ball at the end of its fall.?
Verified Answer
A machine which is 75% efficient uses 12 J of energy in lifting up a 1...
Efficiency = (Output energy / Input energy) x 100
75 = (Output energy / 12 Joule) x 100
Output energy = (75 x 12 Joule) / 100 = 9 Joule

Actual energy utilised to lift the mass up is 9 Joule

Energy required to lift the mass up = mgh
9 J = (1 kg x 10 m/s^2 x h)
h = 0.9 m

Final velocity of freely falling body after falling from a height “H” is
v = sqrt(2gH)
= √(2 x 10 m/s^2 x 0.9 m)
= √(18 m^2/s^2)
= 4.24 m/s

∴ Velocity at the end of its free fall is 4.24 m/s

[Note: I took g = 10 m/s^2 for easy calculations]
This question is part of UPSC exam. View all Class 10 courses
Most Upvoted Answer
A machine which is 75% efficient uses 12 J of energy in lifting up a 1...
Given:
- Efficiency of the machine, η = 75% = 0.75
- Energy used to lift the mass, E = 12 J
- Mass of the ball, m = 1 kg

To find:
The velocity of the ball at the end of its fall.

Solution:

Step 1: Calculate the work done to lift the mass
The work done to lift the mass is given by the formula:
Work = Force × Distance

Since the force required to lift the mass is equal to its weight, we can write:
Work = Weight × Distance

The weight of the mass is given by the formula:
Weight = Mass × Acceleration due to gravity

Substituting the given values:
Weight = 1 kg × 9.8 m/s² = 9.8 N

Therefore, the work done to lift the mass is:
Work = 9.8 N × Distance

Since the work done is equal to the energy used, we have:
9.8 N × Distance = 12 J

Step 2: Calculate the distance
From the equation derived in Step 1, we can rearrange it to solve for the distance:
Distance = 12 J / 9.8 N

Distance = 1.22 m (rounded to two decimal places)

Step 3: Calculate the gravitational potential energy gained
The gravitational potential energy gained when the ball is lifted to a certain height is given by the formula:
Potential Energy = Mass × Acceleration due to gravity × Height

Since the ball is lifted through a distance equal to the height it will fall, the potential energy gained is equal to the energy used to lift the ball:
Potential Energy = 12 J

Step 4: Calculate the kinetic energy at the end of the fall
The efficiency of the machine is given by the formula:
Efficiency = Useful Energy Output / Energy Input

Since the energy used to lift the ball is the input energy, we can write:
Efficiency = Useful Energy Output / 12 J

Solving for Useful Energy Output:
Useful Energy Output = Efficiency × 12 J
Useful Energy Output = 0.75 × 12 J
Useful Energy Output = 9 J

The total mechanical energy at the end of the fall is the sum of the potential energy gained and the useful energy output:
Total Mechanical Energy = Potential Energy + Useful Energy Output
Total Mechanical Energy = 12 J + 9 J
Total Mechanical Energy = 21 J

Step 5: Calculate the velocity at the end of the fall
The total mechanical energy at the end of the fall is equal to the kinetic energy of the ball:
Total Mechanical Energy = (1/2) × Mass × Velocity²

Substituting the given values:
21 J = (1/2) × 1 kg × Velocity²

Simplifying the equation:
42 J = Velocity²

Taking the square root of both sides:
Velocity = √42 m/s

Therefore, the velocity of the ball at the end of its fall is approximately 6.48 m/s (rounded to two decimal places).

Conclusion:
The velocity of the ball at the end of its fall is approximately 6.48 m/s.
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A machine which is 75% efficient uses 12 J of energy in lifting up a 1 kg mass through a certain distance. The mass is allowed to call through that distance. Find the velocity of that ball at the end of its fall.?
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