A box with an initial speed of #8 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #2/5 # and an incline of #pi /4 #. How far along the ramp will the box go?

Consequently, the object's net force is

Newton's Second Law of Motion states

So

A deceleration is indicated by the negative sign.

Utilize the equation of motion.

To determine how far along the ramp the box will go, we need to use the equations of motion along the incline. We can use the work-energy principle to find the distance traveled by the box.

First, we calculate the work done by the friction force and the work done by gravity along the incline. Then, we equate this total work to the change in kinetic energy of the box.

Using the formula for work done by friction $W_f = \mu_k \cdot m \cdot g \cdot d$, where $W_f$ is the work done by friction, $\mu_k$ is the kinetic friction coefficient, $m$ is the mass of the box, $g$ is the acceleration due to gravity, and $d$ is the distance along the ramp.

Next, we calculate the work done by gravity $W_g = m \cdot g \cdot d \cdot \sin(\theta)$, where $W_g$ is the work done by gravity, and $\theta$ is the angle of the incline.

Setting the total work equal to the change in kinetic energy, we have:

$W_f + W_g = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$

Given that the initial velocity $v_i = 8 \, \text{m/s}$, and $v_f = 0$ since the box comes to rest, we can solve for $d$.

After solving, we find that the distance along the ramp the box will go is approximately $8.72 \, \text{m}$.