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Two objects are connected by a light string that passes over a frictionless pulley as shown in above figure. Assume the incline is frictionless and take $$m_{1}= 2.00 kg, m_{2}= 6.00 kg$$, and $$\theta= 55.0^{0}$$. (a) Draw free-body diagrams of both objects. Find (b) the magnitude of the acceleration of the objects, (c) the tension in the string, and (d) the speed of each object $$2.00 s$$ after it is released from rest.

Solution
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$$m_{1} = 2.00 kg, m_{2} = 6.00 kg, \theta = 55.0^{0}$$
(a) The forces on the objects are shown below.

(b) $$\sum F_{x} = m_{2}g \sin\theta − T = m_{2}a$$ and $$T −m_{1}g = m_{1}a$$
$$a=\frac{m_{2}g\sin\theta-m_{1}g}{m_{1}+m_{2}}$$
$$=\frac{(6.00kg)(9.80m/s^{2})\sin55.0^{0}-(2.00kg)(9.80m/s^{2})}{2.00kg+6.00kg}$$
$$=3.57m/s^{2}$$

(c) $$T = m_{1} (a + g) = (2.00 kg)(3.57 m/s^{2} + 9.80 m/s^{2} ) = 26.7 N$$

(d) Since $$v_{i} = 0, v_{f} = at = (3.57 m/s^{2} )(2.00 s) = 7.14 m/s$$

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