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

a).
`W_(a)=5.46x10^-3 J`

b). m=287.78g

Step-by-step 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=(W_(a))/(g*d*(sin(24)-u*cos(24))`

`m=(5.46x^-3J)/(9.81m/s^2*84.0x10^-3*(sin(24)-0.420*cos(24))`

`m=0.2877 kg (1000g)/(1kg)=287.78g`