Question

You are moving a heavy wooden table (596.08kg) from the kitchen to the dining room. The coefficient of friction static for the table and the floor is (Mustatic 0.28) (Mukinetic 0.17). What force must be applied to get the table moving with a constant velocity after the initial motion?

Answer 1

Since table is moving on floor with constant speed

So we can say that it must have net force ZERO

it means that applied force is balance by kinetic friction on the table

So we can say

$F_a = F_k$

here we can say

$F_k = \mu_k F_n$

now we know that

$F_n = mg$

$F_n = 596.08* 9.8$

$F_n = 5841.6 N$

Now we have friction force as

$F_f = 0.17 * 5841.6 = 993.1 N$

So here our applied force is same as kinetic friction

$F = 993.1 N$

Answer 2

when the table moves at constant velocity, kinetic frictional force acts on it. hence the applied must be equal to kinetic frictional force to keep the object moving at constant velocity.

N = normal force on the table by floor

m = mass of the table = 596.08 kg

g = acceleration due to gravity = 9.8 m/s²

μ = coefficient of kinetic friction = 0.17

F = applied force on the table

f = kinetic frictional force on the table

using equilibrium of force in vertical direction

N = mg

N = (596.08) (9.8)

N = 5841.6 N

kinetic frictional force is given as

f = μN

inserting the values

f = (0.17) (5841.6)

f = 993.1 N

along the horizontal direction , force equation is given as

F - f = 0

F = f

F = 993.1 N

hence the applied force comes out to be 993.1 N