Question

A 5 kg block is attached by means of a string to a 2 kg block on a 30 slope. The string is passed over a 3 kg pulley. The co-efficient of sliding friction between the 2 kg block and the slope is 0.2.











Find the velocity of the hanging masspiece after the hanging masspiece has moved down 0.5 m. Solve this problem two different ways:

(a) Using energy considerations.
(b) Using Newton’s Second Law of Motion.

Answer 1

Answer on Question #70794, Physics / Mechanics | Relativity

A 5 kg block is attached by means of a string to a 2 kg block on a $30{}^{\circ}$ slope. The string is passed over a 3 kg pulley. The co-efficient of sliding friction between the 2 kg block and the slope is 0.2. Find the velocity of the hanging mass piece after the hanging mass piece has moved down 0.5 m. Solve this problem two different ways:

(a) Using energy considerations.

(b) Using Newton's Second Law of Motion.

Solution:

a) Initial energy: $M_{5kg}gH$

Final energy: $\frac{M_{5kg}v^2}{2} + m_{2kg}gh + m_{2kg}v_{3kg}^2$

According to energy conservation law: $M_{5kg}gH = \frac{M_{5kg}v^2}{2} + m_{2kg}gh + \frac{m_{2kg}v^2}{2} + A_{friction}$

Geometrically: $h = H * \sin 30{}^\circ$, $A_{friction} = \mu m_{2kg}g * \cos 30{}^\circ * H$

So, final speed is: $v^2 \left( M_{5kg} + m_{2kg} \right) = M_{5kg} g H - m_{2kg} g H \sin 30{}^\circ - \mu m_{2kg} g H * \cos 30{}^\circ$

$$v = \sqrt{ \frac{ M_{5kg} g H - m_{2kg} g H \sin 30{}^\circ - \mu m_{2kg} g * H \cos 30{}^\circ }{ M_{5kg} + m_{2kg} } } = 1.6 \frac{m}{s}$$ $$\text{b)} F = F_{gr_{5kg}} - F_{gr_{2kg}} * \sin 30{}^\circ - F_{friction} * \cos 30{}^\circ$$ $$F = M_{5kg}g - m_{3kg}g * \sin 30{}^\circ - \mu m_{3kg}g * \cos 30{}^\circ = (M_{5kg} + m_{2kg})a$$ $$H = \frac{v^2}{a} \rightarrow v = \sqrt{aH} = \sqrt{ \frac{ M_{5kg} g H - m_{2kg} g H \sin 30{}^\circ - \mu m_{2kg} g * H \cos 30{}^\circ }{ M_{5kg} + m_{2kg} } } = 1.6 \frac{m}{s}$$

Answer: Final speed is $1.6 \, \text{m/s}$

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