A 21kg crate is pushed across the floor at a constant velocity by a 350N force. Determine the coefficient of kinetic friction.
350=mu*21*9.8
solve for mu
M*g = 21 * 9.8 = 205.8 N. = Wt. of crate = Normal force(Fn).
Fk = u*Fn = 205.8u.
Fap-Fk = M*a.
350-205.8u = 21*0,
u = ?.
To determine the coefficient of kinetic friction, we first need to understand the physics concept involved. The force of kinetic friction can be calculated using the equation:
fk = μk * N
where fk is the force of kinetic friction, μk is the coefficient of kinetic friction, and N is the normal force.
In this case, since the crate is being pushed at a constant velocity, we know that the net force acting on the crate is zero. The net force is the vector sum of all the forces acting on the object. In this scenario, the force of friction opposes the applied force, resulting in a net force of zero.
Therefore, the equation can be written as:
Fapplied - fk = 0
Substituting the values given in the problem:
350N - fk = 0
Solving for fk, we can determine the force of kinetic friction.
fk = 350N
Next, we need to determine the normal force (N). The normal force is the force exerted on an object that is in contact with a surface and acts perpendicular to that surface.
In this case, the normal force is equal to the weight of the crate since it is on a horizontal surface. The weight (W) can be calculated using the equation:
W = m * g
where m is the mass of the crate and g is the acceleration due to gravity.
Substituting the given mass:
W = 21kg * 9.8 m/s^2
W = 205.8N
Finally, substitute the values for fk and N into the equation for the force of kinetic friction fk = μk * N:
350N = μk * 205.8N
Dividing both sides of the equation by 205.8N:
μk = 350N / 205.8N
μk ≈ 1.70
The coefficient of kinetic friction is approximately 1.70.