please help with question below?
You need to move a 680 kg log with your tractor. The coefficient of static friction between the log and the ground is µs=0.700, and the coefficient of kinetic friction is µk=0.600. Your tractor applies a horizontal force of magnitude F=5.0 kN to the log. What is the acceleration of the log? You can assume that the gravitational acceleration g=10m/s2.
- oubaasLv 73 weeks ago
once it starts moving :
a = (F-m*g*µs)/m = (5000-680*10*0.7)/680 = 0.35 m/sec^2 (just 2 sign. digits)
once in motion :
a' = (F-m*g*µk)/m = (5000-680*10*0.6)/680 = 1.4 m/sec^2 (just 2 sign. digits)
- NCSLv 73 weeks ago
First, you must determine if the static friction force is overcome.
max fs = µs*m*g = 0.700 * 680kg * 10m/s² = 4760 N
which is < 5000 N applied, so yes.
a = net F / mass = (5000 - 0.600*680kg*10m/s²) / 680kg
a = 1.35 m/s²
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