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
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 is#0# ) 
#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"" "))#