Afootball player at practice pushes a 60 kg blocking sled across the field at a constant speed. the coefficient of kinetic friction between the grass and the sled is 0.30. how much force must he apply to the sled?

180N

Using Newton's law of motion;

∑F = m x a       --------------------(i)

Where;

∑F = Resultant force

m = mass of the object (sled in this case)

a = acceleration of the sled

Calculate the resultant force;

Since the direction of motion is horizontal, the horizontal forces acting on the sled are the;

i. Applied force () in one direction and;

ii. Frictional force () in the other direction to oppose motion

Therefore, the resultant force ∑F is the vector sum of the two forces. i.e;

∑F = -  -----------------------(i)

Frictional force is the product of the coefficient of kinetic friction (μ) and weight(W) of the sled. i.e

= μ x W

Where;

W = mass(m) x gravity(g)

W = m x g

=> = μmg

Substitute into equation (ii)

∑F = - μmg

Substitute ∑F into equation (i)

- μmg = ma  -------------------(iii)

Since the motion is at constant speed, it means acceleration is zero (0)

Substitute a = 0 into equation (iii) to give;

- μmg = 0

=> = μmg

Substitute the values of μ = 0.3, m = 60kg and g = 10m/s² into the above equation to give;

=>  = 0.3 x 60 x 10

=> = 180N

This means that the applied force should be 180N

anchor = archer //