A body slipping on a rough horizontal plane moves with a deceleration of 4.0m/{s^2} What is the coefficient of kinetic friction between the block and the plane?

To find the coefficient of kinetic friction (μk\mu_k) between the body and the rough horizontal plane, we can use Newton's second law and the formula for the frictional force.
Step-by-step solution:
1. Identify the forces involved:
o The body is experiencing a deceleration, which is due to the frictional force acting opposite to the motion.
o The frictional force (fkf_k) is given by: fk=μkNf_k = \mu_k N where:
 μk\mu_k is the coefficient of kinetic friction,
 NN is the normal force, which for a horizontal surface is equal to the weight of the object: N=mgN = mg, where mm is the mass of the object, and gg is the acceleration due to gravity (9.8 m/s29.8 \, \text{m/s}^2).
2. Use Newton's second law:
o The net force acting on the object is causing the deceleration. According to Newton’s second law, the net force FnetF_{\text{net}} is also equal to the mass times the acceleration: Fnet=maF_{\text{net}} = ma
o Since the only horizontal force is the frictional force, the frictional force is responsible for the deceleration. Therefore, we have: fk=maf_k = ma where aa is the deceleration of the body, given as 4.0 m/s24.0 \, \text{m/s}^2.
3. Set up the equation:
o Substitute the expression for friction fk=μkmgf_k = \mu_k mg into the equation fk=maf_k = ma: μkmg=ma\mu_k mg = ma
o Simplifying this equation: μkg=a\mu_k g = a μk=ag\mu_k = \frac{a}{g}
4. Substitute the known values:
o The deceleration a=4.0 m/s2a = 4.0 \, \text{m/s}^2,
o The acceleration due to gravity g=9.8 m/s2g = 9.8 \, \text{m/s}^2.
So,
μk=4.09.8\mu_k = \frac{4.0}{9.8}
5. Calculate the result:
μk≈0.408\mu_k \approx 0.408
Final answer:
The coefficient of kinetic friction between the block and the plane is approximately 0.41.