A gorilla drags a log across the jungle floor. The log begins at rest and accelerates at until the gorilla is moving at a fast lope. The coefficient of kinetic friction between the log and the mud and plants of the jungle is . Assume that friction and the gorillas pulling force remain constant the whole time. What frictional force acts on the log

  fr = 245 N,  T = fr

Explanation:

We can solve this exercise using the equilibrium conditions, let's start by creating a reference system

where we call T the drag force of the gorilla, m the mass of the trunk and μ the coefficient of kinetic friction

Y axis  

           N- W =

           N = W = mg

X axis

          T - fr = 0

          T = fr

the expression for the friction force is

          fr = μ N

          fr = μ mg

we substitute

           T = μ m g

When analyzing these expressions we see that when the trunk reaches the maximum speed that the gorilla carries, the friction force is equal to the gorilla's tension

            fr = T

for a specific calculation we must assign values to:

      μ = 0.25

      m = 100 kg

let's calculate

         T = 0.25 100 9.8

         T = 245 N

therefore for this case

         fr = 245 N

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2