Yes the relationship between them is direct linear ..
Complete Question
A 95 kg clock initially at rest on a horizontal floor requires a 650 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 560 N keeps it moving with a constant velocity. Find the coefficient of static friction and the coefficient of kinetic friction.
Answer:
The value for static friction is 
The value for static friction is 
Explanation:
From the question we are told that
The mass of the clock is 
The first horizontal force is 
The second horizontal force is 
Generally the static frictional force is equal to the first horizontal force
So

=> 
=> 
Generally the kinetic frictional force is equal to the second horizontal force
So



Answer:
1.196 μm
Explanation:
D = Screen distance = 3 m
= Wavelength = 598 m
y = Distance of first-order bright fringe from the center of the central bright fringe = 4.84 mm
d = Slit distance


For first dark fringe

Wavelength of first-order dark fringe observed at this same point on the screen is 1.196 μm
Answer:
2.52 m/s
Explanation:
When the man takes a step, his foot is stationary while his body revolves around it. At the point when his body is directly above his foot, there will be no normal force at his maximum speed.
Sum of the forces in the radial direction:
∑F = ma
mg = m v² / r
g = v² / r
v = √(gr)
Given that r = 0.650 m:
v = √(9.8 m/s² × 0.650 m)
v = 2.52 m/s