A 15 kg box is given an initial push so that it slides across the floor & comes to a stop. If the coefficient of friction is .30,
A) Find the Ffr Fg=mg=Fg=15(9.8)=147N
Ffr=mu(Fn)- Ffr=.3*147N=44.1N
B)Find the acceleration. Hint: what is the net force as the box slides to a stop?
I want to say 0 but then C) asks what d will be if it's initial speed is 3.0 m/s & I can't get any answer other than 0 if a & V. are 0...
the force of friction is 15*9.8*.3
acceleration= forcefriction/mass
B) To find the acceleration, we can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration (Fnet = ma). In this case, the net force is the force of friction, Ffr, which opposes the motion of the box.
Since the box comes to a stop, the net force must be equal to zero. Therefore, we can set the equation Ffr = ma to zero and solve for acceleration:
Ffr = ma
44.1N = 15kg * a
a = 44.1N / 15kg
a = 2.94 m/s^2
So, the acceleration of the box as it slides and comes to a stop is approximately 2.94 m/s^2.
C) To find the distance, d, when the initial speed, V, is 3.0 m/s, we can use the equation of motion:
V^2 = V0^2 + 2ad
Where:
V = final velocity (0 m/s since it comes to a stop)
V0 = initial velocity (3.0 m/s)
a = acceleration (2.94 m/s^2)
d = distance
Plug in the values and solve for d:
0^2 = (3.0m/s)^2 + 2 * (2.94m/s^2) * d
0 = 9.0m^2/s^2 + 5.88m/s^2 * d
-9.0m^2/s^2 = 5.88m/s^2 * d
d = -9.0m^2/s^2 / 5.88m/s^2
d = -1.53m
The negative sign indicates that the distance is in the opposite direction of the initial motion. So, the distance traveled by the box before coming to a stop is approximately 1.53 meters in the opposite direction of the initial velocity.
To find the acceleration of the box, we need to consider the net force acting on it. The net force is the vector sum of all the forces acting on an object. In this case, the only force acting on the box is the friction force (Ffr). Since the box is coming to a stop, this friction force opposes the motion and acts in the direction opposite to the initial motion.
The net force is given by the equation:
net force = mass × acceleration
Since the box is coming to a stop, the net force is equal to zero:
0 = 15 kg × acceleration
Therefore, the acceleration of the box is 0 m/s².
Now, let's move on to part C. The distance (d) covered by an object can be determined by the equation:
d = (initial velocity × time) + (1/2 × acceleration × time²)
In this case, the initial velocity is 3.0 m/s, the acceleration is 0 m/s² (as we found in part B), and we need to find the time it takes for the box to come to a stop.
To find the time, we can use the equation:
final velocity = initial velocity + (acceleration × time)
Since we know that the final velocity is 0 m/s (as the box comes to a stop), we can set up the equation:
0 = 3.0 m/s + (0 m/s² × time)
Simplifying the equation, we find:
3.0 m/s = 0
This equation is not possible, meaning the box will not come to a stop if its initial velocity is 3.0 m/s. Therefore, the distance traveled (d) cannot be determined in this scenario.