Question

A.) A hardworking ant must supply 0.0650 N to pull a piece of fruit at constant velocity 8.40 cm up the colony's ant hill. If the coefficient of kinetic friction between the piece of fruit and the 24.0° sloped ant hill is 0.420, calculate the work done by the ant by pulling the piece of fruit up the hill.
B.) Use the work-energy theorem to calculate the mass of the piece of fruit in grams. (The acceleration due to gravity is 9.81 m/s2.)

Answer 1

Answer:

a). $W_{a}=5.46x10^-3 J$

b). m=287.78g

Explanation:

a)

The work done by the hardworking ant is the same work in the direction of the plane so

$W_{a}=F*d$

$W_{a}=0.0650N*84.0x10^-3m$

$W_{a}=5.46x10^-3 J$

b)

Using the theorem of work energy and conservation can find the mass of the piece of fruit

$W_{a}=E_{k}+E_{p}+W_{fk}$

$E_{k}=0$

$E_{p}=m*g*h$

$W_{k}=u*N$

$W_{a}=0+m*g*d*sin(24)-u*m*g*d*cos(24)$

$W_{a}=m*g*d*(sin(24)-u*cos(24))$

Resolve to m

$m=\frac{W_{a}}{g*d*(sin(24)-u*cos(24)}$

$m=\frac{5.46x^-3J}{9.81m/s^2*84.0x10^-3*(sin(24)-0.420*cos(24)}$

$m=0.2877 kg \frac{1000g}{1kg}=287.78g$