Question

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

Answer 1

fr = 245 N, T = fr

Step-by-step 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