A 0.5 kg object is given an initial velocity of 3 m/s after which it slides a distance of 8 m across a level floor. What is the coefficient of kinetic friction between the object and the floor?
M*g = 0.5 * 9.8 = 4.9 N. = Wt. of object = Normal force(Fn).
V^2 = Vo^2 + 2a*d.
0 = 9 + 2a*8,
a = -0.563 m/s^2.
Fk = u*Fn = u*4.9 = 4.9u = Force of kinetic friction.
Fap-Fk = M*a.
0 - 4.9u = 0.5*(-0.563),
u = 0.057.
acceleration is not 10 since it is no a free fall, i think alfred can find alternative solving to attemp in order to have more convenient answer
average speed = (3+0)/2 = 1.5 m/s
so t = 8/1.5 = 80/15 = 5.33 seconds sliding to a stop
then acceleration a = (0-3)/5.33 = -.5625 m/s^2
so now the physics
F = m a
-m g (mu) = m (-.5625)
mu = .5625/9.81 = .0573
Well, to determine the coefficient of kinetic friction, we need to use the equation of motion for the object. But before that, let me tell you a joke about friction to lighten the mood:
Why don't scientists trust atoms?
Because they make up everything!
Now, getting back to the question, we know that the work done against friction is equal to the initial kinetic energy of the object. So, we can use the work-energy principle:
Work done against friction (W) = Initial kinetic energy (0.5 * mass * velocity^2)
Now we can write the equation:
W = (0.5 * 0.5 kg*(3 m/s)^2)
Simplifying the equation, we get:
W = 2.25 Joules
Since work done against friction can be calculated using the equation W = force of friction * distance, we get:
Force of friction * distance = 2.25 Joules
Since the distance is given as 8 m, we can divide both sides of the equation by 8:
Force of friction = 2.25 Joules / 8 m
Simplifying, we find:
Force of friction = 0.28125 Newtons
The frictional force can also be written as the coefficient of kinetic friction multiplied by normal force. The normal force is given by the mass multiplied by the acceleration due to gravity (9.8 m/s^2).
So, we can rewrite the equation as:
Coefficient of kinetic friction * mass * acceleration due to gravity = 0.28125 Newtons
Plugging in the values, we have:
Coefficient of kinetic friction * 0.5 kg * 9.8 m/s² = 0.28125 N
Simplifying further, we find:
Coefficient of kinetic friction = 0.28125 N / (0.5 kg * 9.8 m/s²)
After doing the math,
Coefficient of kinetic friction ≈ 0.057
So the coefficient of kinetic friction between the object and the floor is approximately 0.057.
To find the coefficient of kinetic friction between the object and the floor, we can use the equation:
F_k = μ_k * N
where:
F_k is the force of kinetic friction
μ_k is the coefficient of kinetic friction
N is the normal force
The normal force is the force exerted by the floor perpendicular to the surface of the object. On a level floor, the normal force is equal to the weight of the object.
First, let's find the weight of the object:
Weight (W) = mass (m) * acceleration due to gravity (g)
where:
m = 0.5 kg (mass of the object)
g = 9.8 m/s^2 (acceleration due to gravity)
W = 0.5 kg * 9.8 m/s^2
W = 4.9 N
Now, we need to find the force of kinetic friction (F_k) by using the equation:
F_k = mass (m) * acceleration (a)
Since the object is sliding, the only force opposing its motion is the force of kinetic friction. Therefore, the force of kinetic friction is equal to the force that caused the object to decelerate:
F_k = m * a
Using the equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s, since the object comes to a stop)
u = initial velocity (3 m/s)
a = acceleration
s = distance (8 m)
0^2 = 3^2 + 2 * a * 8
0 = 9 + 16a
16a = -9
a = -9/16
a = -0.5625 m/s^2
Now, we can find the force of kinetic friction:
F_k = m * a
F_k = 0.5 kg * (-0.5625 m/s^2)
F_k = -0.28125 N
Note: The negative sign indicates that the force of kinetic friction acts in the opposite direction to the object's motion.
Finally, we can substitute the force of kinetic friction (F_k) and weight (W) into the equation F_k = μ_k * N:
-0.28125 N = μ_k * 4.9 N
Now, we can solve for the coefficient of kinetic friction (μ_k):
μ_k = -0.28125 N / 4.9 N
μ_k = -0.057
The coefficient of kinetic friction between the object and the floor is approximately -0.057.
from energy=
1/2mv^2.
=1/2×0.5×3^2
=2.25 but
work done=force applied
then force=2.25
recall f=ma
=0.5×10
5N
so.
coefficient=kinetic frictions limiting friction
2.25÷5=0.45
coefficient of kinetic friction =0.45