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ludmilkaskok [199]
3 years ago
10

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 th

e 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.)
Physics
1 answer:
barxatty [35]3 years ago
5 0

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

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