Question

An athlete needs to lose weight and decides to do it by "pumping iron." (a) How many times must an 60.0 kg weight be lifted a distance of 0.670 m in order to burn off 1.00 lb of fat, assuming that that much fat is equivalent to 3500 kcal? (b) If the weight is lifted once every 1.80 s, how long does the task take?

Answer 1

Answer:

a) 37171

b) $20650.5$ sec

Explanation:

m = mass of the weight being lifted = 60.0 kg

d = distance by which the weight is lifted = 0.670 m

E = Energy available to burn = 1 lb = 3500 kcal = 3500 x 4184 J

g = acceleration due to gravity = 9.8 m/s²

n = number of times the weight is lifted

Energy available to burn is given as

E = n m g d

3500 x 4184 = n (60) (9.8) (0.670)

n = 37171

b)

T = time period for each lift up = 1.80 s

t = total time taken

Total time taken is given as

$t = \frac{n}{T}$

$t = \frac{37171}{1.80}$

$t = 20650.5$ sec