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

Apr 28, 2016

$l = 2 , 41 \text{ meters}$

Explanation: $\text{total energy ":E_k=1/2*m*v^2" (at A)}$

$\text{energy changed heat due to friction :} {W}_{f} = \mu \cdot m \cdot g \cdot \cos \left(\frac{\pi}{3}\right) \cdot l$

$\text{potential energy for the point of A :} {E}_{p} = m \cdot g \cdot h$

$\text{we can write the energy conservation equation as :}$

${E}_{k} = {E}_{p} + {W}_{f}$

$\frac{1}{2} \cdot m \cdot {v}^{2} = m \cdot g \cdot h + \mu \cdot m \cdot g \cdot \cos \left(\frac{\pi}{3}\right) \cdot l$

$\text{so ;} h = l . \sin \left(\frac{\pi}{3}\right)$

$\frac{1}{2} \cdot \cancel{m} \cdot {v}^{2} = \cancel{m} \cdot g \cdot l \cdot \sin \left(\frac{\pi}{3}\right) + \mu \cdot \cancel{m} \cdot g \cdot \cos \left(\frac{\pi}{3}\right) \cdot l$

$\frac{1}{2} \cdot {v}^{2} = g \cdot l \cdot 0 , 87 + \frac{1}{3} \cdot g \cdot 0 , 5 \cdot l$

${v}^{2} = g \cdot l \left(2 \cdot 0 , 87 + \frac{2 \cdot 0 , 5}{3}\right)$

$v = 7 \frac{m}{s}$

${7}^{2} = g \cdot l \left(1 , 74 + \frac{1}{3}\right)$

$49 = g \cdot l \left(\frac{5 , 22 + 1}{3}\right)$

$49 = g \cdot l \cdot 2 , 07$

$g = 9 , 81 \frac{m}{s} ^ 2$

$49 = 9 , 81 \cdot l \cdot 2 , 07$

$49 = 20 , 31 \cdot l$

$l = 2 , 41 \text{ meters}$