Question31. (III) A block is placed on a table as shown in Fig. 5-41. The coefficient of kinetic friction between the block and the table is 0.35 . The block is connected by massless ropes over pulleys (whose mass and friction can be ignored) to a block on the right, and a block on the left, as shown in Fig. 5-41. Find the acceleration of the block on the table if it is moving to the right.
FIGURE 5-41
Problem 31.
Studdy Solution
STEP 1
1. The block on the table has a mass of .
2. The coefficient of kinetic friction between the block and the table is .
3. The block on the right has a mass of .
4. The block on the left has a mass of .
5. The ropes are massless and the pulleys are frictionless.
6. The system is moving to the right.
7. We need to find the acceleration of the block on the table.
STEP 2
1. Analyze the forces acting on each block.
2. Write the equations of motion for each block.
3. Solve the system of equations to find the acceleration.
STEP 3
Analyze the forces acting on each block.
- For the block on the table:
- Force of gravity:
- Normal force:
- Frictional force:
- Tension in the rope due to block:
- Tension in the rope due to block:
- For the block:
- Force of gravity:
- Tension in the rope:
- For the block:
- Force of gravity:
- Tension in the rope:
STEP 4
Write the equations of motion for each block.
- For the block on the table:
$ T_1 - T_2 - F_f = 2.5 \cdot a
\]
- For the block:
$ 5.0 \cdot 9.8 - T_1 = 5.0 \cdot a
\]
- For the block:
$ T_2 - 3.0 \cdot 9.8 = 3.0 \cdot a
\]
STEP 5
Solve the system of equations to find the acceleration.
1. Calculate the frictional force:
$ F_f = 0.35 \cdot (2.5 \cdot 9.8) = 8.575 \, \text{N}
\]
2. Substitute equations for and from the and blocks into the equation for the block:
$ (5.0 \cdot 9.8 - 5.0 \cdot a) - (3.0 \cdot 9.8 + 3.0 \cdot a) - 8.575 = 2.5 \cdot a
\]
3. Simplify and solve for :
49 - 5a - 29.4 - 3a - 8.575 = 2.5a
\]
11.025 = 10.5a
\]
$ a = \frac{11.025}{10.5} \approx 1.05 \, \text{m/s}^2
\]
The acceleration of the block on the table is approximately .
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