A box with an initial speed of #7 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #7/3 # and an incline of #(5 pi )/12 #. How far along the ramp will the box go?

Question

A box with an initial speed of #7  m/s# is moving up a ramp.  The ramp has a kinetic friction coefficient of #7/3 # and an incline of #(5  pi )/12 #.  How far along the ramp will the box go?

Nathan Ashcroft · Accepted Answer

The distance is #=1.06m#

Taking the direction up and parallel to the plane as positive #↗^+#

The coefficient of kinetic friction is #mu_k=F_r/N#

Consequently, the object's net force is

#F=-F_r-Wsintheta#

#=-F_r-mgsintheta#

#=-mu_kN-mgsintheta#

#=mmu_kgcostheta-mgsintheta#

Newton's Second Law states

#F=m*a#

Where #a# is the acceleration

So

#ma=-mu_kgcostheta-mgsintheta#

#a=-g(mu_kcostheta+sintheta)#

The coefficient of kinetic friction is #mu_k=7/3#

The incline of the ramp is #theta=5/12pi#

#a=-9.8*(7/3cos(5/12pi)+sin(5/12pi))#

#=-23.08ms^-2#

A deceleration is indicated by the negative sign.

We utilize the equation of motion.

#v^2=u^2+2as#

#u=7ms^-1#

#v=0#

#a=-23.08ms^-2#

#s=(v^2-u^2)/(2a)#

#=(0-49)/(-2*23.08)#

#=1.06m#

Nathan Ashcroft · Answer

undefined To find how far along the ramp the box will go, we can use the equations of motion and the principles of physics. First, we'll calculate the acceleration of the box up the ramp using the components of the gravitational force and the frictional force. Then, we'll use the kinematic equation to find the distance traveled.

The acceleration up the ramp can be found using the following equation:

$$ a = g \sin(\theta) - \mu_k g \cos(\theta) $$

where $ g $ is the acceleration due to gravity, $ \theta $ is the angle of the incline, and $ \mu_k $ is the coefficient of kinetic friction.

Given:
- $ g = 9.8 \, \text{m/s}^2 $ (acceleration due to gravity)
- $ \theta = \frac{5\pi}{12} $
- $ \mu_k = \frac{7}{3} $

Substituting the values, we get:

$$ a = (9.8 \, \text{m/s}^2) \sin\left(\frac{5\pi}{12}\right) - \left(\frac{7}{3}\right)(9.8 \, \text{m/s}^2) \cos\left(\frac{5\pi}{12}\right) $$

Then we find:

$$ a ≈ -5.978 \, \text{m/s}^2 $$

The negative sign indicates that the acceleration is acting in the opposite direction to the motion of the box.

Now, we can use the equation of motion to find the distance traveled:

$$ d = v_i t + \frac{1}{2} a t^2 $$

Where:
- $ v_i $ is the initial velocity (7 m/s)
- $ t $ is the time
- $ a $ is the acceleration

Since the box eventually comes to rest, its final velocity $ v_f = 0 $. We can use this information to find the time taken ($ t $):

$$ v_f = v_i + at $$
$$ 0 = 7 - 5.978t $$
$$ t ≈ 1.17 \, \text{s} $$

Now, plug this time value into the equation to find the distance traveled:

$$ d = (7 \, \text{m/s})(1.17 \, \text{s}) + \frac{1}{2}(-5.978 \, \text{m/s}^2)(1.17 \, \text{s})^2 $$
$$ d ≈ 5.56 \, \text{m} $$

So, the box will travel approximately 5.56 meters along the ramp before coming to rest.