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.)

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davidlcastiblanco davidlcastiblanco · Answer 1 · 2019-11-08T17:16:48+01:00

Answer:

a). [tex]W_{a}=5.46x10^-3 J[/tex]

b). m=287.78g

Explanation:

a)

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

[tex]W_{a}=F*d[/tex]

[tex]W_{a}=0.0650N*84.0x10^-3m[/tex]

[tex]W_{a}=5.46x10^-3 J[/tex]

b)

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

[tex]W_{a}=E_{k}+E_{p}+W_{fk}[/tex]

[tex]E_{k}=0[/tex]

[tex]E_{p}=m*g*h[/tex]

[tex]W_{k}=u*N[/tex]

[tex]W_{a}=0+m*g*d*sin(24)-u*m*g*d*cos(24)[/tex]

[tex]W_{a}=m*g*d*(sin(24)-u*cos(24))[/tex]

Resolve to m

[tex]m=\frac{W_{a}}{g*d*(sin(24)-u*cos(24)}[/tex]

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

[tex]m=0.2877 kg \frac{1000g}{1kg}=287.78g[/tex]