A box with an initial speed of $4 \frac{m}{s}$ $4 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{1}{3}$ $1/3$ and an incline of $\frac{2 π}{3}$ $(2 pi )/3$ . How far along the ramp will the box go?

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Answer 1 · 2016-06-14T12:22:04

$Delta x = (4 sqrt[3])/g$ [m]

The initial speed is $v_0$ so the initial kinetic energy is given by

$E_k = 1/2mv_0^2$

The box friction work loses along the ramp are given by

$mu xx mg xx sin(theta) Delta x$

with $mu$ the kinetic friction coefficient, $theta$ the ramp incline, and $Delta x$ the covered distance along the ramp.

So we can equate

$1/2mv_0^2 = mu xx mg xx sin(theta) Delta x + m g sin(theta) Delta x$

Solving for $Delta x$

$Delta x =(1/2 v_0^2)/((mu+1) sin(theta)g ) = (4 sqrt[3])/g$ [m]

A box with an initial speed of 4 m/s4ms4 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 1/3 131/3 and an incline of (2 pi )/3 2π3(2 pi )/3 . How far along the ramp will the box go?