A box with an initial speed of $7 \frac{m}{s}$ $7 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{1}{3}$ $1/3$ and an incline of $\frac{π}{3}$ $( pi )/3$ . How far along the ramp will the box go?

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A box with an initial speed of $7 \frac{m}{s}$ $7 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{1}{3}$ $1/3$ and an incline of $\frac{π}{3}$ $( pi )/3$ . How far along the ramp will the box go?

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Answer 1 · 2016-04-28T19:13:30

$l = 2, 41 meters$ $l=2,41 " meters"$

Explanation:

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$total energy : E_{k} = \frac{1}{2} \cdot m \cdot v^{2} (at A)$ $"total energy ":E_k=1/2*m*v^2" (at A)"$

$energy changed heat due to friction : W_{f} = μ \cdot m \cdot g \cdot cos (\frac{π}{3}) \cdot l$ $"energy changed heat due to friction :" W_f=mu* m*g*cos (pi/3)*l$

$potential energy for the point of A : E_{p} = m \cdot g \cdot h$ $"potential energy for the point of A :"E_p=m*g*h$

$we can write the energy conservation equation as :$ $"we can write the energy conservation equation as :"$

$E_{k} = E_{p} + W_{f}$ $E_k=E_p+W_f$

$\frac{1}{2} \cdot m \cdot v^{2} = m \cdot g \cdot h + μ \cdot m \cdot g \cdot cos (\frac{π}{3}) \cdot l$ $1/2*m*v^2=m*g*h+mu*m*g*cos (pi/3)*l$

$so ; h = l . sin (\frac{π}{3})$ $"so ;"h=l.sin (pi/3)$

$1/2*cancel(m)*v^2=cancel(m)*g*l*sin(pi/3)+mu*cancel(m)*g*cos (pi/3)*l$

$1/2*v^2=g*l*0,87+1/3*g*0,5*l$

$v^2=g*l(2*0,87+(2*0,5)/3)$

$v=7 m/s$

$7^2=g*l(1,74+1/3)$

$49=g*l((5,22+1)/3)$

$49=g*l*2,07$

$g=9,81 m/s^2$

$49=9,81*l*2,07$

$49=20,31*l$

$l=2,41 " meters"$

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A box with an initial speed of $7 \frac{m}{s}$ $7 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{1}{3}$ $1/3$ and an incline of $\frac{π}{3}$ $( pi )/3$ . How far along the ramp will the box go?

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A box with an initial speed of $7 \frac{m}{s}$ $7 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{1}{3}$ $1/3$ and an incline of $\frac{π}{3}$ $( pi )/3$ . How far along the ramp will the box go?