An object with a mass of # 14 kg# is lying still on a surface and is compressing a horizontal spring by #70 c m#. If the spring's constant is # 2 (kg)/s^2#, what is the minimum value of the surface's coefficient of static friction?

Question

An object with a mass of # 14  kg# is lying still on a surface and is compressing a horizontal spring by #70  c   m#. If the spring's constant is # 2  (kg)/s^2#, what is the minimum value of the surface's coefficient of static friction?

Isabella Alvarez · Accepted Answer

The coefficient of static friction is #=0.01#

The mass is #m=14kg#

The compression of the spring is #x=0.7m#

The spring constant is #k=2kgs^-2#

The reaction of the spring is #R=kx=2*0.7=1.4N#

The acceleration due to gravity is #g=9.8ms^-2#

The normal reaction of the object is

#N=mg=14*9.8=137.2N#

The coefficient of static friction is

#mu_s=R/N=1.4/137.2=0.01#

Leo Abeyta · Answer

undefined The minimum value of the surface's coefficient of static friction is 0.98.

Nora Arzate · Answer

undefined The minimum value of the surface's coefficient of static friction can be calculated using the formula:

$ \mu_s = \frac{kx}{mg} $

where:
- $ \mu_s $ is the coefficient of static friction,
- $ k $ is the spring constant (2 (kg)/s$^2$),
- $ x $ is the compression of the spring (70 cm or 0.7 m),
- $ m $ is the mass of the object (14 kg),
- $ g $ is the acceleration due to gravity (9.8 m/s$^2$).

Substituting the given values into the formula:

$ \mu_s = \frac{(2 \, \text{kg/s}^2) \times (0.7 \, \text{m})}{(14 \, \text{kg}) \times (9.8 \, \text{m/s}^2)} $

$ \mu_s = \frac{1.4}{137.2} $

$ \mu_s \approx 0.0102 $

Therefore, the minimum value of the surface's coefficient of static friction is approximately 0.0102.