Question

The 20 kg suitcase A is released from rest at point C of height 3 m from floor ED. The suitcase slides down a friction-less ramp between C and E and continues sliding on the floor until stopping at a certain distance from E.
The coefficient of friction between the floor ED and suitcase A is µ(ED) = 0.4.
Consider g = 9.8 m/s2
Calculate the velocity reached in m/s at point E of suitcase A

Answer 1

Answer on Question#49048 – Engineering – Other

$$\sqrt{58.8} \approx 7.668 \frac{m}{s}$$

Solution

Consider suitcase $A$ is not rotating. Thus, as soon as ramp $CE$ – frictionless, we can write next conservation law: $mg(h_{1} - h_{0}) + \frac{mV_{1}^{2}}{2} = mg(h_{2} - h_{0}) + \frac{mV_{2}^{2}}{2}$, where $m$ – mass of suitcase. $h, V$ – It’s height and velocity respectively. $h_{0}$ – ground level. Note: there is no influence of ramp’s angle, the only thing that matter is initial height.

Consider: $g = 9.8\frac{m}{s^2}$; height at point $E \equiv h_2 \equiv h_0 = 0$; Initial velocity of suitcase $V_{1} = 0$; $h_1 \equiv h$; $V_{2} \equiv V$.

Hence,

$$mgh = \frac{mV^{2}}{2}; \quad gh = \frac{V^{2}}{2}; \quad V = \sqrt{2gh}$$

$h$ – height at point $C$. $V$ – velocity at point $E$.

$$V = \sqrt{2 * 9.8 * 3} = \sqrt{58.8} \frac{m}{s}$$

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