Question

A student eats a dinner rated as 2000 calories. He wishes to do an equivalent amount of work in the gymnasium by lifting a 50 kg mass. How many times must he raise the weight to expend this much energy? Assume he raised the weight a distance 2 m each time and that no work is done when the weight is dropped to the floor. 1 food calorie = 10 ^ 3 * cal 1cal = 4.186​

Answer 1

By using work energy theorem

Intake energy = work out

Intake energy = (number of lifting barbell) x (work done per lifting)

Ei = n x W

Ei = nW

Now

Work per lifting = Force x displacement

W = (mg) x (displacement)

W = (50 x 9.8) x (2)

W = 980 J

Now

Ei = n x W (Ei = 2000 Cal = 2 kilocalory = 8368 Joule)

8368 = n x 980

n = 8.53

Aprrox 8 times he should raise barbell to expend 2000 calory.

Step-by-step explanation: