The drawings show three identical blocks sliding along horizontal surfaces. Each block has a different mass, as indicated. The coefficient of kinetic friction in each case is the same. Which block has the greatest deceleration?

A = 1m B = 3m C = 2m

a. A
b. B
c. C
d. The deceleration is the same in each case.

Option D

Explanation:

Given:

- Three different masses on a horizontal surface slowing down due to kinetic friction,

Find:

- Deceleration for each case

Solution:

- We will take an arbitrary case of mass m moving on a flat surface, with coefficient of kinetic friction as u_k.

- We will use Newton's second law of motion in the horizontal direction:

                                         F_net = m*a

- Where the only external force on the block is the frictional force F_f.

                                         F_f = m*a

- We know that F_f is related to the normal contact force of the body. We will apply equilibrium conditions on the block in vertical direction:

                                         F_n - m*g = 0

                                         F_n = m*g

And,                                  F_f = u_k*F_n

Hence,                             F_f = u_k*m*g

- Now plug in the expression of F_f in to our first equation of motion:

                                         u_k*m*g = m*a

Simplify,

                                          a = u_k*g

- We see that a body with any arbitrary mass with the conditions given is independent of the mass of the object. The coefficient of kinetic friction and gravitational acceleration are constants. Hence, all three masses will decelerate the same. Hence, Option D

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.