The law of conservation of energy and energy efficiency!!!!!!!

Can you please answer the following questions, show clear working out, and please make sure the answer is correct, thank you:

1. If a school athlete of mass 55kg raises the centre of his body 0.8m from the ground, what is his increase in gravitational potential energy? What Kinetic energy will be needed to do this? With what speed will he need to leave the ground?

2. Given a power input of 400 watts, calculate the efficiency of devices that produce outputs of:

a) 80 watts

b) 300 watts

3. A catapult stores 25j of elastic potential energy, but the projectile that fires has a kinetic energy of only 15j. What is the catapults efficiency?

4. If the electric motor of a forklift has an efficiency of 90% how much energy is needed to increase a crates gravitational potential energy by 2000j?

5. In a torch, the efficiency of a battery in converting chemical to electrical energy is 85%, and the efficiency of the globe in converting electrical into light energy is only 2%. What is the overall efficiency?

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1) The potential energy gained =mgh, where m =mass 55 kg, g is gravitational acceleration=9.81m/s^2 not given but taken from tascientific tables, h = 0.8m = (1/2) mv^2 the kinetic energy because of the speed v with whivc he jumps.

So v^2 = 2gh Or v =sqrt( 2*9.81*0.8) = 3.9618m/s

2)

Efficiency = (out power/input power)100

a)Efficiency = (80/400)100=25%

b) Efficiency = (300/400)100 = 75%

3) Efficiency = (Output KE/Input PE)100 = (15/25)100 = 60%

4) Efficiency = (Output energy/Input energy)100. Output energy =2000j efficiency = 90%. Therefore,

Input energy = Output energy/efficiency = 2000J/90%

= (2000//90)100 = 2222.22 Joule.

5)Over all efficiency = (Efficiency of converting from chemical to electrical energy)* (efficiency of conversion from electrical energy to light energy) = (85%)(2%) = (0.85*0.02)100 = 1.7%

1. Kinetic energy will be needed raise body and speed when leaving the ground?

When the body of mass m = 55 kg is raised by height h = 0.8 m. The increase in potential energy (e) is given by the formula:

e = m*h*g

Where g = acceleration due to gravity = 9.81 m/s^2

Applying value for given variables in the above formula:

e = 55*0.8*9.81 = 431.64 J

Kinetic energy of an object is given by the formula:

e = (1/2)*m*(v^2)

When v represent the velocity required to enable the athlete to attain the the potential energy this kinetic energy will be equal to the potential energy:

Thus:

431.64 = (1/2)*55*(v^2)

Therefore:

v^2 = 431.64*2/55 = 15.696

And:

v = 3.962 m/s

2. Given a power input of 400 watts, calculate the efficiency of devices that produce outputs of:

a) 80 watts

Efficiency = (80/400)*100 = 20%

b) 300 watts

Efficiency = (300/400)*100 = 75%

3. A catapult stores 25j of elastic potential energy, but the projectile that fires has a kinetic energy of only 15j. What is the catapults efficiency?

Efficiency = (15/25)*100 = 60%

4. If the electric motor of a forklift has an efficiency of 90% how much energy is needed to increase a crates gravitational potential energy by 2000j?

Energy needed = (Potential energy)*100/Efficiency

= 2000*100/90 = 2222.2222 J

5. In a torch, the efficiency of a battery in converting chemical to electrical energy is 85%, and the efficiency of the globe in converting electrical into light energy is only 2%. What is the overall efficiency?

Overall efficiency = (Chemical efficiency)*(Globe efficiency/100

= 85*2/100 = 1.7%

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