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 distance in m from point E reached by suitcase A as it stops.

Answer 1

Answer on Question#49528 - Engineering - Other

The $m = 20$ kg suitcase $A$ is released from rest at point $C$ of height $h = 3\mathrm{m}$ 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 $\mu(ED) = 0.4$ . Consider $g = 9.8\frac{\mathrm{m}}{\mathrm{s}^2}$ .

Calculate the distance in m from point E reached by suitcase A as it stops.

Solution:

Since the ramp is friction less, according to the law of conservation of energy the energy of suitcase at points $C$ and $E$ are equal. While sliding on the floor the suitcase loses its kinetic energy to overcome the friction force. The work needed to overcome this force on the interval $ED$ is

$$W = F _ {f} \cdot E D$$

The friction force $F_{f}$ can be easily found, since we know the coefficient of friction $\mu(ED)$ and the normal force $F_{N}$ on the interval $ED$ , which is equal to the weight $m \cdot g$ of the suitcase. So the friction force is

$$F _ {f} = F _ {N} \cdot \mu (E D) = m \cdot g \cdot \mu (E D)$$

and the work needed to overcome it is

$$W = m \cdot g \cdot \mu (E D) \cdot E D$$

The kinetic energy of the suitcase at the point $E$ equals the potential energy difference of the suitcase between points $C$ and $E$ , which equals

$$U _ {C E} = m \cdot g \cdot h$$

To find the distance $ED$ we need to equate this energy to $W$ :

$$m \cdot g \cdot \mu (E D) \cdot E D = m \cdot g \cdot h$$

And finally

$$ED = \frac{h}{\mu(ED)} = \frac{3\,\mathrm{m}}{0.4} = 7.5\,\mathrm{m}$$

Answer: $ED = \frac{h}{\mu(ED)} = 7.5\,\mathrm{m}$.

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