Question

A 56 g Frisbee is thrown from a point 1.3 m above the ground with a speed of 15 m/s. When it has reached a height of 2.1 m, its speed is 12 m/s. What was the reduction in the mechanical energy of the Frisbee-Earth system because of air drag

Answer 1

Answer:

The mechanical energy of frisbee-earth system because of air is equal to $-2.2232J$

Explanation:

From work energy theorem ;the total work done is equal to change in its kinetic energy

$w_{g}+w_{air}=K.E_{f}-K.E_{i}$

$w_{g}$ = work done by gravity = $mg(h_{1}-h_{2})$  

$w_{air}$ = work done by air drag

    From above equations

$w_{air}$ = $\frac{56}{1000\times 2 } \times(12^{2} -15^{2}) +\frac{56}{1000} \times(2.1-1.3)J$

$w_{air}=-2.2232J$