When you push a 1.82-{\rm kg} book resting on a tabletop, it takes 2.21 N to start the book sliding. Once it is sliding, however, it takes only 1.52 N to keep the book moving with constant speed.
1-What is the coefficient of static friction between the book and the tabletop?
2-What is the coefficient of kinetic friction between the book and the tabletop?
forcefriction=pushing force when..
mu*mg=2.21 solve for mu, then do it again for kinetic coefficent.
To find the coefficient of static friction between the book and the tabletop, we can use the equation:
Fs = μs * N
Where:
Fs is the force of static friction
μs is the coefficient of static friction
N is the normal force
The normal force N can be calculated as:
N = m * g
Where:
m is the mass of the book
g is the acceleration due to gravity (approximately 9.8 m/s^2)
Given:
m = 1.82 kg
Plugging in the values:
N = 1.82 kg * 9.8 m/s^2
N ≈ 17.836 N
Now, we can use the equation for static friction to find the coefficient of static friction:
Fs = μs * N
Given:
Fs = 2.21 N
Plugging in the values:
2.21 N = μs * 17.836 N
Now, solving for the coefficient of static friction:
μs = 2.21 N / 17.836 N
μs ≈ 0.124
Therefore, the coefficient of static friction between the book and the tabletop is approximately 0.124.
To find the coefficient of kinetic friction between the book and the tabletop, we can use the equation:
Fk = μk * N
Where:
Fk is the force of kinetic friction
μk is the coefficient of kinetic friction
N is the normal force (same as before)
Given:
Fk = 1.52 N
Plugging in the values:
1.52 N = μk * 17.836 N
Now, solving for the coefficient of kinetic friction:
μk = 1.52 N / 17.836 N
μk ≈ 0.085
Therefore, the coefficient of kinetic friction between the book and the tabletop is approximately 0.085.