Find the coefficient of kinetic friction?

Hi Sun,

First we want to determine the acceleration of the block.

We start with the equation:

v^2 = u^2 + 2as

where v = final velocity, u = initial velocity (in this case 0), a = acceleration and s = distance.

Therefore,

1^2 = 0^2 + 2*a*1.4

Therefore,

1 = 2.8*a

Therefore,

a = 1/2.8 = 0.36m/s^2

Now that we have the acceleration of the block, we need to find the net force on the block.

Net Force = Mass*Acceleration

= 40*0.36

= 14.4 Newtons

Next we need to determine the force of friction. We use the equation:

Net Force = Applied Force - Force of friction

Therefore,

14.4 = 85 - Force of friction

Therefore,

Force of friction = 85 - 14.4

= 70.6 Newtons

Now to obtain the coefficient of kinetic friction we use the equation that you have listed:

uk = fk/FN

Here FN is the "normal force", which is the force between the block and floor due to gravity. This is given by:

FN = m*g = 40*9.8 = 392 Newtons.

fk is the force of friction which we determined to be 70.6 Newtons.

Therefore, uk = 70.6/392 = 0.18

let F is the force applied

F = 85N

m = 40kg

as the object is initially at rest , so

u = 0 , v = 1 m/s , a=? , s = 1.4 m

v^2 - u^2 = 2as

1 - 0 = 2*a*1.4

1 = 2.8*a

a = 0.35m/s^2

as the friction always opposes motion so it is in opposite direction to the applied force

F(net) = F - fk

ma = 85 - fk

fk = 85 - 40*0.35

= 71N

also

fk = ukN (N = normal reaction)

= uk * mg

71 = uk * 40*9.8

uk = 71/40*9.8 = 0.18N

Hi,

Have you drawn you diagram with outline (perhaps dotted) of box in initial position along with your outline of box in it's final position at displacement 1.4m then drawn forces acting on the box at the initial position?