A 5.00-kg box slides 7.00 m across the floor before coming to rest. What is the coefficient of kinetic friction between the floor and the box if the box had an initial speed of 3.00 m/s?
Initial speed = 3 m/s
acceleration constant during braking
therefore average speed during stop = 3/2 = 1.5 m/s
time to stop = t =7 m / 1.5 m/s = 4.67 seconds
v = Vi +a t
0 = 3 + a(4.67)
a = -0.643 m/s^2
F = m a
F = 5 (.643)
F = 3.21 Newtons
F = mu (weight)
3.21 = mu (5)(9.81)
mu = - .0655
thank you.
You are welcome. As always, check my arithmetic.
Nonsensical
To find the coefficient of kinetic friction (μk) between the floor and the box, we'll need to use the formulas of kinematic motion and force.
First, let's find the deceleration (a) of the box using the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, as the box comes to rest)
u = initial velocity (3.00 m/s)
a = deceleration
s = distance traveled (7.00 m)
Rearranging the equation, we get:
a = (v^2 - u^2) / (2s)
a = (0 - (3.00)^2) / (2 * 7.00)
Now, we can calculate the deceleration:
a = -1.29 m/s^2
The negative sign indicates that the acceleration is in the opposite direction of motion (opposing the motion).
Next, we can calculate the net force (Fnet) acting on the box using Newton's second law of motion:
Fnet = ma
Where:
m = mass of the box (5.00 kg)
a = deceleration (-1.29 m/s^2)
Substituting the values, we get:
Fnet = (5.00 kg) * (-1.29 m/s^2)
Fnet = -6.45 N
The negative sign indicates that the net force is also in the opposite direction of motion (opposing the motion).
Finally, we can determine the frictional force (Ff) using the equation:
Ff = μk * FN
Where:
μk = coefficient of kinetic friction (what we're trying to find)
FN = normal force
Since the box is at rest, the normal force (FN) is equal to the weight of the box (mg), where g is the acceleration due to gravity (9.8 m/s^2):
FN = mg
FN = (5.00 kg) * (9.8 m/s^2)
FN = 49 N
Now, we can calculate the coefficient of kinetic friction:
Ff = μk * FN
-6.45 N = μk * 49 N
Simplifying the equation:
μk = -6.45 N / 49 N
From this calculation, we find that the coefficient of kinetic friction between the floor and the box is approximately -0.132 (rounded to three decimal places). Note that the negative sign indicates the direction of the frictional force opposing the motion of the box.