A box with an initial speed of 3 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 5/7 and an incline of ( 5 pi )/12 . How far along the ramp will the box go?
1 Answer
Explanation:
We're asked to find how far the box will travel up the ramp, given its initial speed, the coefficient of kinetic friction, and the angle of inclination.
I will solve this problem using only Newton's laws and kinematics (i.e. without using work/energy).
I will take the positive
x -direction as *up the ramp.*
When the box reaches its maximum distance, the instantaneous velocity will be
ul((v_x)^2 = (v_(0x))^2 + 2a_x(Deltax)
where
-
v_x is the instantaneous velocity (which is0 ) -
v_(0x) is the initial velocity -
a_x is the (constant) acceleration -
Deltax is the distance it travels (what we're trying to find)
Since the velocity
0 = (v_(0x))^2 + 2a_x(Deltax)
We also figure that the acceleration will be negative because it slows down and comes to a brief stop at its maximum height, so we then have
0 = (v_(0x))^2 + 2(-a_x)(Deltax)
And we can move it to the other side:
ul(2a_x(Deltax) = (v_(0x))^2
Rearranging for the distance traveled
ulbar(|stackrel(" ")(" "Deltax = ((v_(0x))^2)/(2a_x)" ")|)
We already know the initial velocity, so we need to find the acceleration of the box.
Let's use Newton's second law of motion to find the acceleration, which is
ul(sumF_x = ma_x
where
-
sumF_x is the net force acting on the box -
m is the mass of the box -
a_x is the acceleration of the box (what we're trying to find)
The only forces acting on the box are
-
the gravitational force (acting down the ramp), equal to
mgsintheta -
the friction force (acting down the ramp), equal to
mu_kn
And so we have our net force equation:
sumF_x = mgsintheta + mu_kn The normal force
n exerted by the incline is equal to
n = color(purple)(mgcostheta) So we can plug this in to the net force equation above:
ul(sumF_x = mgsintheta + mu_kcolor(purple)(mgcostheta)
Or
ul(sumF_x = mg(sintheta + mu_kcostheta)
Now, we can plug this in for
sumF_x = ma_x
ma_x = mg(sintheta + mu_kcostheta)
We can cancel the mass
color(green)(ul(a_x = g(sintheta + mu_kcostheta)
Now that we have found an expression for the acceleration, let's plug it into the equation
Deltax = ((v_(0x))^2)/(2a_x)
And we get
color(red)(ulbar(|stackrel(" ")(" "Deltax = ((v_(0x))^2)/(2g(sintheta + mu_kcostheta))" ")|)
We're given in the problem
-
v_(0x) = 3 "m/s" -
theta = (5pi)/12 -
mu_k = 5/7 -
and the gravitational acceleration
g = 9.81 "m/s"
Plugging these in:
color(blue)(Deltax) = ((3color(white)(l)"m/s")^2)/(2(9.81color(white)(l)"m/s"^2)(sin[(5pi)/12] + 5/7cos[(pi)/12])) = color(blue)(ulbar(|stackrel(" ")(" "0.399color(white)(l)"m"" ")|)