Question

Really need help with this Physics question!

Which applied force will allow a 25 newton block of steel to continue sliding on a steel table? The block of steel has a kinetic coefficient of friction of 0.57 and a static coefficient of friction of 0.74.

A) 3.5 N
B) 7.0 N
C) 14.0 N
D) 18.0 N

Answer 1

As we know that block of steel is continue to be in moving state

so here the friction must be kinetic friction between two surface

so we know that formula of kinetic friction must be

`F_k = \mu_k mg`

now we have

`\mu_k = 0.57`

`mg = 25 N`

now from the above equation we have

`F_k = 0.57* 25`

`F_k = 14.25 N`

So here we need atleast 14.25 N force to continue sliding the box now as per given options all forces which are less than 14.25 N is not correct

Hence correct answer must be 18.0 N

D) 18.0 N