Question

A person pushes a 500kg crate with a force of 700 N across a concrete floor. Given the coefficicent of kinetic friction is 0.100 what is the acceleration of the crate across the floor?

Answer 1

We know that kinetic friction is the opposing force that is set up between the surfaces of contact between the two bodies, when one body is in relative motion over the surface of the other body.

If coefficient of kinetic friction is given by u;

Then u = F/R

Where F is the kinetic friction and R is the normal reaction.  

Re arranging the above eqn,we get

F= uR ....... Eqn 1

Also, once the body has started sliding over the other, acceleration a produced in body would be

a = (F' - F)/M ......eqn 2

Where M is the mass of the body and F' is the external force applied on it

Using eqn 1 in 2; we get

a= (F' - uR)/M ....eqn 3

Given, F' = 700 N

M= 500 kg

u = 0.100

Also, F = uR.

Further,R= mg where g is the acceleration due to gravity,= 9.8 m/s​​​​​​2​​​ . Using these value in expression for F, we get

i.e F = 0.100*500*9.8 = 490

Using value of F,M,F' in eqn 3, we get

a = 700-490/500

= 210/500

a = 0.42 m/s​​​​​​2​​​​​ ... Acceleration produced in the crate.