Question

A 51.0 kg cheetah accelerates from rest to its top speed of 31.7 m/s. HINT (a) How much net work (in J) is required for the cheetah to reach its top speed? J (b) One food Calorie equals 4186 J. How many Calories of net work are required for the cheetah to reach its top speed? Note: Due to inefficiencies in converting chemical energy to mechanical energy, the amount calculated here is only a fraction of the energy that must be produced by the cheetah's body. Cal

Answer 1

(a) $2.56\cdot 10^4 J$

The work-energy theorem states that the work done on the cheetah is equal to its change in kinetic energy:

$W= \Delta K = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

where

m = 51.0 kg is the mass of the cheetah

u = 0 is the initial speed of the cheetah (zero because it starts from rest)

u = 31.7 m/s is the final speed

Substituting, we find

$W=\frac{1}{2}(51.0 kg)(31.7 m/s)^2 - \frac{1}{2}(51.0 kg)(0)^2=2.56\cdot 10^4 J$

(b) 6.1 cal

The conversion between calories and Joules is

1 cal = 4186 J

Here the energy the cheetah needs is

$E=2.56\cdot 10^4 J$

Therefore we can set up a simple proportion

$1 cal : 4186 J = x : 2.56\cdot 10^4 J$

to find the equivalent energy in calories:

$x=\frac{(1 cal)(2.56\cdot 10^4 J)}{4186 J}=6.1 cal$