Two blocks, A and B, are connected by a light string over a frictionless pulley. Block B is on a rough incline. The masses of A and B are 10 kg and 5.0 kg, respectively. The slope... Two blocks, A and B, are connected by a light string over a frictionless pulley. Block B is on a rough incline. The masses of A and B are 10 kg and 5.0 kg, respectively. The slope angle is 30 degrees, and the coefficient of kinetic friction between block B and the slope is 0.30. Initially, the blocks are released from rest with the string taut. When block A has fallen 40 cm, find: (a) the change in gravitational potential energy of block A, (b) the change in gravitational potential energy of block B, (c) the kinetic frictional force acting on block B, (d) the gain in thermal energy in the system, and (e) the speeds of blocks A and B at that instant. Read more

Steps to Solve

1. Calculate the change in gravitational potential energy of block A

The change in gravitational potential energy ($ \Delta U_A $) is given by $ \Delta U_A = -m_A g \Delta h $, where $m_A$ is the mass of block A, $g$ is the acceleration due to gravity (9.8 m/s$^2$), and $\Delta h$ is the distance block A falls.

$$ \Delta U_A = - (10 \text{ kg}) (9.8 \text{ m/s}^2)(0.40 \text{ m}) = -39.2 \text{ J} $$

2. Calculate the change in gravitational potential energy of block B

The distance block B rises along the incline is the same as the distance block A falls, which is 0.40 m. The vertical height block B rises is $ \Delta h_B = \Delta h \sin(\theta)$, where $\theta$ is the angle of the incline. The change in gravitational potential energy of block B ($ \Delta U_B $) is given by $ \Delta U_B = m_B g \Delta h_B = m_B g \Delta h \sin(\theta) $.

$$ \Delta U_B = (5.0 \text{ kg}) (9.8 \text{ m/s}^2) (0.40 \text{ m}) \sin(30^\circ) = (5.0)(9.8)(0.40)(0.5) = 9.8 \text{ J} $$

3. Calculate the kinetic frictional force acting on block B

The kinetic frictional force ($f_k$) is given by $f_k = \mu_k N$, where $\mu_k$ is the coefficient of kinetic friction and $N$ is the normal force. The normal force is $N = m_B g \cos(\theta)$. Therefore, $f_k = \mu_k m_B g \cos(\theta)$.

$$ f_k = (0.30)(5.0 \text{ kg})(9.8 \text{ m/s}^2) \cos(30^\circ) = (0.30)(5.0)(9.8)(\sqrt{3}/2) \approx 12.73 \text{ N} $$

4. Calculate the gain in thermal energy in the system

The gain in thermal energy ($E_{thermal}$) is equal to the work done by friction, which is $E_{thermal} = f_k \Delta h $.

$$ E_{thermal} = (12.73 \text{ N})(0.40 \text{ m}) = 5.092 \text{ J} \approx 5.09 \text{ J} $$

5. Calculate the speeds of blocks A and B

The total mechanical energy change is $ \Delta E = \Delta U_A + \Delta U_B + \Delta K + E_{thermal} = 0 $. Since the blocks start from rest, the change in kinetic energy is $\Delta K = \frac{1}{2} m_A v^2 + \frac{1}{2} m_B v^2 = \frac{1}{2} (m_A + m_B)v^2$. Therefore, $0 = \Delta U_A + \Delta U_B - E_{thermal} + \frac{1}{2}(m_A + m_B)v^2$ $ - \frac{1}{2}(m_A + m_B)v^2 = \Delta U_A + \Delta U_B - E_{thermal} $ Multiply both sides by -1 $ \frac{1}{2}(m_A + m_B)v^2 = - \Delta U_A - \Delta U_B + E_{thermal} $

$$ \frac{1}{2} (10 \text{ kg} + 5.0 \text{ kg}) v^2 = -(-39.2 \text{ J}) - 9.8 \text{ J} - 5.092 \text{ J} $$

$$ \frac{1}{2} (15 \text{ kg}) v^2 = 39.2 \text{ J} - 9.8 \text{ J} - 5.092 \text{ J} = 24.308 \text{ J} $$

$$ v^2 = \frac{2(24.308 \text{ J})}{15 \text{ kg}} = \frac{48.616}{15} \approx 3.241 $$

$$ v = \sqrt{3.241} \approx 1.80 \text{ m/s} $$