Question

A football 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?

Answer 1

Answer:

The force must he apply to the sled is of F= 764.4 N.

Explanation:

m= 60 kg

g= 9.8 m/s²

μ=0.3

W= m*g

W= 588 N

Fr= μ*W

Fr= 176.4 N

F= W + Fr

F= 764.4 N

Answer 2

Answer:

180N

Explanation:

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 ($F_{A}$) in one direction and;

ii. Frictional force ($F_{R}$) in the other direction to oppose motion

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

∑F = $F_{A}$ - $F_{R}$  -----------------------(i)

Frictional force $F_{R}$ is the product of the coefficient of kinetic friction (μ) and weight(W) of the sled. i.e

$F_{R}$ = μ x W

Where;

W = mass(m) x gravity(g)

W = m x g

=> $F_{R}$ = μmg

Substitute $F_{R}$ into equation (ii)

∑F = $F_{A}$ - μmg

Substitute ∑F into equation (i)

$F_{A}$ - μmg = ma  -------------------(iii)

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

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

$F_{A}$ - μmg = 0

=> $F_{A}$ = μmg

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

=> $F_{A}$  = 0.3 x 60 x 10

=> $F_{A}$ = 180N

This means that the applied force should be 180N