A block of mass 40 kg resting on a horizontal surface just begins to s

1.	A block of mass 40 kg resting on a horizontal surface just begins to slide when a horizontal force of 120 N is applied to it. Once the motion starts, It can be maintained by a force of 80 N. Determine the coefficients of static friction and kinetic friction(g = 9.8 m/s2)
Answer» Given: F_L = 120N, F_k = 80N, m = 40 kg, g = 9.8 m/s² To find: i. Coefficient of static friction (μ_S) ii. Coefficient of kinetic friction (μ_k) Formulae: i. μ_s = \(\frac{F_L}{N}\) ii. μ_k = \(\frac{F_K}{N}\) Calculation: From formula (i) we get. N = mg = 40 × 9.8 = 392 N ∴ μ_s = \(\frac{F_L}{N}=\frac{120}{392}\) = 0.306 From formula (ii) we get, ∴ μk = \(\frac{F_k}{N}=\frac{80}{392}\) = 0.204 1. The coefficient of static friction is 0.306. 2. The coefficient of kinetic friction is 0.204.

A block of mass 40 kg resting on a horizontal surface just begins to slide when a horizontal force of 120 N is applied to it. Once the motion starts, It can be maintained by a force of 80 N. Determine the coefficients of static friction and kinetic friction(g = 9.8 m/s2)

Answer»

Given: F_L = 120N, F_k = 80N, m = 40 kg, g = 9.8 m/s²

To find:

i. Coefficient of static friction (μ_S)

ii. Coefficient of kinetic friction (μ_k)

Formulae:

i. μ_s = \(\frac{F_L}{N}\)

ii. μ_k = \(\frac{F_K}{N}\)

Calculation:

From formula (i) we get.

N = mg = 40 × 9.8 = 392 N

∴ μ_s = \(\frac{F_L}{N}=\frac{120}{392}\) = 0.306

From formula (ii) we get,

∴ μk = \(\frac{F_k}{N}=\frac{80}{392}\) = 0.204

1. The coefficient of static friction is 0.306.

2. The coefficient of kinetic friction is 0.204.