A 15kg box is given an initial push so that it slides across the floor & comes to a stop. I the coefficient of friction is .30,
a) find Ffr- Fg=mg - Fg=15kg(9.8N/kg)=147N
Ffr=mu(Fn) - Ffr=.30(147N)=44.1
b) Find the acceleration of the box. Hint: what is the net force as the box slides to a stop?
I really want to say 0 but then c) asks if the initial speed is 3.0 m/s what is d & I'm having a hard time getting anything but 0 w/ the info I have & a=0
In the vertical direction mg down from gravity = mg up from the floor so the vertical acceleration = 0
In the horizontal direction there is only one force, friction, deaccerating
-0.30 m g = m a
so
a = -0.30 g = -0.30 * 9.8 =-2.94 m/s^3
Now if the initial speed is 3
then
v = 3 - a t
v = 3 - 2.94 t
at stop
v = 0
so
0 = 3 - 2.94 t
t = 1.02 seconds to stop
now you can use the distance equation to get distance d
d = 3 t - (1/2)(2.94)t
but it is easier to use average speed = 3/2 = 1.5 m/s for 1.02 sec
1.5 * 1.02 = 1.53 meters :)
v = 3 -
and a =
b) The net force acting on the box is the force of friction, which is given by Ffr = 44.1 N. According to Newton's second law, the net force is equal to the mass of the object multiplied by its acceleration. Therefore, we have:
Fnet = ma
Since the box comes to a stop, its final velocity is zero meaning its acceleration is also zero. Hence, the net force is also zero. Therefore:
44.1 N = ma
Since a = 0, there is no acceleration.
c) If the initial speed of the box is 3.0 m/s, and there is no acceleration, the distance traveled can be calculated using the formula:
d = v * t
Since there is no acceleration, the time taken for the box to come to a stop will depend on the deceleration caused by friction. The formula to find time using deceleration is:
v = u - at
u is the initial velocity, a is acceleration (which is zero in this case), and t is time. Rearranging the formula to solve for t, we get:
t = (u - v) / a
Plugging in the values:
t = (3.0 m/s - 0) / 0
Since a is zero, we end up with division by zero, which is undefined. Therefore, with the given information, it is not possible to determine the distance traveled by the box.
To find the acceleration of the box, we need to consider the net force acting on it. The net force is the sum of all forces acting on the box.
In this case, the only force acting on the box is the friction force (Ffr), given by Ffr = μ(Fn), where μ is the coefficient of friction and Fn is the normal force acting on the box.
Since the box is on a horizontal surface, the normal force is equal to the weight of the box, which is equal to the gravitational force (Fg), given by Fg = mg, where m is the mass of the box and g is the acceleration due to gravity (approximately 9.8 m/s^2).
a) We have already calculated the value of Fg to be 147 N. Now, we can calculate the friction force (Ffr) using the given coefficient of friction (μ = 0.30):
Ffr = μ(Fn) = 0.30(147 N) = 44.1 N
b) Since the box is sliding across the floor and comes to a stop, we know that the direction of the net force is opposite to the direction of the motion. Therefore, the net force can be expressed as:
Net force = -Ffr
As the box comes to a stop, the net force causes deceleration, which is opposite to the initial direction of motion. Thus, the acceleration of the box is negative.
c) If the initial speed of the box is given as 3.0 m/s, and the acceleration is 0, we can determine the distance traveled by using the formula:
v^2 = u^2 + 2as
Where:
v = final velocity (which is 0 m/s since the box comes to a stop)
u = initial velocity (3.0 m/s)
a = acceleration (0)
s = distance traveled (what we are trying to find)
Rearranging the equation gives us:
s = (v^2 - u^2) / (2a)
Substituting the given values:
s = (0^2 - 3.0^2) / (2 * 0)
s = (-9.0) / (0)
Division by zero is undefined, so in this case, we cannot determine the distance traveled (d) with the provided information.