A force of 75.0 N in the horizontal direction is required to set a 37.3-kg box, initially at rest on a flat surface, in motion. After the box begins to move, a horizontal force of 51.3 N is required to maintain a constant speed.
(a) What is the coefficient of static friction between the box and the surface?
(b) What is the coefficient of kinetic friction between the box and the surface?
(a) 75.0 / weight ... 75.0 / (37.3 * g)
(b) 51.3 / weight ... 51.3 / (37.3 * g)
M*g = 37.3 * 9.8 = 365.5 N. = Wt. of box = Normal(Fn).
a. Fs = u*Fn = 365.5u. = Force of static friction.
Fap-Fs = M*a.
75 - 365.5u = M*0.
u = 0.205. = coefficient of static friction.
b. Fap-Fk = M*a.
51.3 - 365.5u = M*0,
u = 0.140. = coefficient of kinetic friction.
Note. Since velocity is constant, acceleration is zero.
To find the coefficient of static friction (μs), we can use the equation:
μs = F_s / N
where F_s is the force of static friction and N is the normal force.
In the given scenario, the normal force (N) is equal to the weight of the box (mg), where m is the mass of the box and g is the acceleration due to gravity (9.8 m/s^2). So, N = 37.3 kg × 9.8 m/s^2 = 365.54 N.
To find the force of static friction (F_s), we know that it is equal to the applied force required to set the box in motion, which is 75.0 N.
So, μs = 75.0 N / 365.54 N ≈ 0.205
Therefore, the coefficient of static friction between the box and the surface is approximately 0.205.
To find the coefficient of kinetic friction (μk), we can use the equation:
μk = F_k / N
where F_k is the force of kinetic friction.
In the given scenario, the force of kinetic friction (F_k) is the force required to maintain a constant speed once the box is in motion, which is 51.3 N.
So, μk = 51.3 N / 365.54 N ≈ 0.140
Therefore, the coefficient of kinetic friction between the box and the surface is approximately 0.140.