PLEASEEE HELPPP!!!
1 answer:
Given :
A mover slides a refrigerator weighing 650 N at a constant velocity across the floor a distance of 8.1 m.
The force of friction between the refrigerator and the floor is 230 N.
To Find :
How much work has been performed by the mover on the refrigerator.
Solution :
Since, refrigerator is moving with constant velocity.
So, force applied by the mover is also 230 N ( equal to force of friction ).
Now, work done in order to move the refrigerator is :
Hence, this is the required solution.
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Answer:
a) The student must run flight of stairs to lose 1.00 kg of fat 709.5 times.
b) Average power
P(w)= 1062.07 [w]
P(hp)=1.42 [hp]
c) This activity is highly unpractical, because the high amount of repetitions he has to due in order to lose, just 1 Kg of fat.
Explanation:
First, lets consider the required amount of work to move the mass of the student. (considering running stairs just as a vertical movement)
Work:
Where m is the mass of the student, g is gravity (9.8 m/s) and d is the total distance going up the stairs (0.15m *85steps= 12.75m )
Converting from Joules to Kcals:
Now lets take into account the efficiency of the human body (20%)
2.537 ---> 20%
x ---> 100%
So the student is consuming 12.685 KCals each time he runs up the stairs.
Now,
1 g --> 9 Kcals
1000 g --> 9000KCals
Burning 1 g of fat, requieres 9 KCals, 1000g burns 9000KCals. So in order to burn a 1Kg of fat:
He must run up the stairs 709.5 times, to burn 1 Kg of fat.
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For b) just converting units, taking into account the time lapse. (53103.75 is the 100% of the energy in joules, from converting 12.685Kcals to joules)
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