A box with an initial speed of 7 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 3/2  and an incline of pi /4 . How far along the ramp will the box go?

Nov 23, 2017

$\sqrt{2}$meters

Explanation:

We know that the initial velocity ( ${v}_{i}$) is $7 m {s}^{-} 1$ and the box will

move on the ramp which makes an angle $\frac{\pi}{4}$ with the horizontal till the velocity becomes zero .

Sp that the final velocity ( ${v}_{f}$ ) of the box is zero.

A kinematic relation connecting intial ,final velocity and the displacement(S) is given by

${v}_{f}^{2} - {v}_{i}^{2} = 2 \cdot a \cdot S$

where

a is the acceleration of the box ..

So the only thing we need to know is the acceleration of the box.

acceleration is the ratio of force acting on the box along the direction of motion of the box to the mass of the box.

$a = \frac{F}{M}$

where F is the force and M is the mass of the box.

THERE ARE ONLY OPPOSING FORCES ACTING ON THE BOX.

They are
1. force due to gravity.
2. frictional force.

Not the complete gravitational force acting on the box along it's motion only the negative of the vertical component acts along the direction of the motion that is $- M g \cdot \sin \left(\frac{\pi}{4}\right)$

The frictional force $- \mu M g \cos \left(\frac{\pi}{4}\right)$

So that the net force acting on it will be

sum of gravitational and frictional force

$F = - \frac{M g}{\sqrt{2}} \left(1 + \mu\right)$

where $\mu$ is the coefficient of kinetic friction ..

There fore acceleration is

$a = \frac{\cancel{M} g \left(1 + \mu\right)}{\cancel{M} \sqrt{2}}$

SUBSTITUTE THE VALUES TO GET ACCELERATION:

$a = \frac{9.8 \cdot \left(1 + \frac{3}{2}\right)}{\sqrt{2}} \implies \frac{5 \times 9.8}{2 \sqrt{2}}$

Back onto displacement relation

${v}_{i} = 7 m {s}^{-} 1$
${v}_{f} = 0 m {s}^{-} 1$
${7}^{2} - 0 = 2 \cdot a \cdot s$

$\frac{49}{2 a} = S$

$S = \frac{49 \times 2 \sqrt{2}}{98} \implies \sqrt{2} m$