Questions: A block with mass m1 hangs from a rope that is extended over an ideal pulley and attached to a second block with mass m2 that sits on a ledge slanted at an angle of 20° (see figure below). Suppose the system of blocks is initially held motionless and, when released, begins to accelerate. (a) If m1=7.25 kg, m2=3.00 kg, and the magnitude of the acceleration of the blocks is 0.170 m / s^2, find the magnitude of the kinetic frictional force between the second block and the ledge. (b) What is the value of the coefficient of kinetic friction between the block and the ledge?

A block with mass m1 hangs from a rope that is extended over an ideal pulley and attached to a second block with mass m2 that sits on a ledge slanted at an angle of 20° (see figure below). Suppose the system of blocks is initially held motionless and, when released, begins to accelerate.
(a) If m1=7.25 kg, m2=3.00 kg, and the magnitude of the acceleration of the blocks is 0.170 m / s^2, find the magnitude of the kinetic frictional force between the second block and the ledge.

(b) What is the value of the coefficient of kinetic friction between the block and the ledge?

Transcript text: A block with mass $m_{1}$ hangs from a rope that is extended over an ideal pulley and attached to a second block with mass $m_{2}$ that sits on a ledge slanted at an angle of $20^{\circ}$ (see figure below). Suppose the system of blocks is initially held motionless and, when released, begins to accelerate. (a) If $m_{1}=7.25 \mathrm{~kg}, m_{2}=3.00 \mathrm{~kg}$, and the magnitude of the acceleration of the blocks is $0.170 \mathrm{~m} / \mathrm{s}^{2}$, find the magnitude of the kinetic frictional force between the second block and the ledge. N (b) What is the value of the coefficient of kinetic friction between the block and the ledge?

Solution

Solution Steps

Step 1: Identify the forces acting on each block

For block $ m_1 $:
- Gravitational force: $ F_{g1} = m_1 g $
- Tension in the rope: $ T $
For block $ m_2 $:
- Gravitational force: $ F_{g2} = m_2 g $
- Normal force: $ N $
- Frictional force: $ f_k $
- Tension in the rope: $ T $

Step 2: Write the equations of motion for each block

For block $ m_1 $ (moving vertically): \[ m_1 g - T = m_1 a \]
For block $ m_2 $ (moving along the incline): \[ T - m_2 g \sin(20^\circ) - f_k = m_2 a \]

Step 3: Solve for the tension $ T $

From the equation for $ m_1 $: \[ T = m_1 g - m_1 a \] Substituting the given values: \[ T = 7.25 \times 9.8 - 7.25 \times 0.17 = 71.05 - 1.2325 = 69.8175 \, \text{N} \]

Step 4: Solve for the frictional force $ f_k $

Substitute $ T $ into the equation for $ m_2 $: \[ 69.8175 - 3 \times 9.8 \times \sin(20^\circ) - f_k = 3 \times 0.17 \] \[ 69.8175 - 10.053 - f_k = 0.51 \] \[ f_k = 69.8175 - 10.053 - 0.51 = 59.2545 \, \text{N} \]