Math

QuestionQuestion 2 A 2.55 kg box, starting from rest, slides down a 5.10 m long ramp inclined at 29.029.0^{\circ} from the horizontal. Halfway down the ramp, it hits a second box, with a mass of 1.65 kg . The two boxes stick together and slide the rest of the way down the ramp. The coefficient of kinetic friction between the ramp and the boxes is 0.200. How fast are they going when they reach the bottom? \square

Studdy Solution
Determine the final velocity of the combined mass at the bottom of the ramp using the kinematic equation:
vfinal2=vcombined2+2acombinedsremaining v_{\text{final}}^2 = v_{\text{combined}}^2 + 2a_{\text{combined}}s_{\text{remaining}}
where sremaining=5.102m s_{\text{remaining}} = \frac{5.10}{2} \, \text{m} .
vfinal=vcombined2+2acombined5.102 v_{\text{final}} = \sqrt{v_{\text{combined}}^2 + 2 \cdot a_{\text{combined}} \cdot \frac{5.10}{2}}
The final velocity of the boxes when they reach the bottom is:
vfinal \boxed{v_{\text{final}}}

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