A dockworker loading crates on a ship finds that a 26-kg crate, initially at rest on a horizontal surface, requires a 72-N horiz
ontal force to set it in motion. However, after the crate is in motion, a horizontal force of 54 N is required to keep it moving with a constant speed. Find the coefficients of static and kinetic friction between crate and floor. g
A 26-kg crate, initially at rest on a horizontal surface, requires a 72-N horizontal force to set it in motion.
Maximum Frictional force on stationary object F = μs R where μs is coefficient of static friction and R is reaction of ground which is equal to weight for object placed on horizontal surface .
F = μs x mg
Given F = 72 , m = 26
72 = 26 x 9.8 x μs
μs = .28
Kinetic friction : ---
Frictional force on moving object F = μk R where μk is coefficient of kinetic friction and R is reaction of ground which is equal to weight for object placed on horizontal surface .
The formula for potential energy would be PE = mgh
Where: m is the mass; g is the acceleration due to gravity, and h would be height.
So plugging in the data:
PE = 150kg × 9.8 m/s^2 × 10m
= 14700 Joules would be the potential energy for this problem.
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Applying the summation of torques about the wedge for equilibrium, taking the clockwise direction as negative. Since the ruler is balanced horizontally about the wedge. Therefore, the summation of all torques acting about the wedge must be equal to zero.