Final answer:
Block B on a horizontal tabletop is attached by very light horizontal strings to two hanging blocks, A and C. 1. The tension on block A is 68.6 N. 2. The tension on block C is 98 N. 3. The magnitude of the acceleration of block B is 1.64 m/s², and its direction is toward the right.
Step-by-step explanation:
1. The tension on block A can be determined by considering the forces acting on block A.
Since the pulleys are ideal, the tension in the string connected to block A is the same as the tension in the string connected to block B.
Block A is being pulled upward by the weight of block B, so the tension in the string is equal to the weight of block B.
Therefore, the tension on block A is 7.00 kg * 9.8 m/s² = 68.6 N.
2. The tension on block C can be determined by considering the forces acting on block C.
Since the pulleys are ideal, the tension in the string connected to block C is the same as the tension in the string connected to block B.
Block C is being pulled downward by its own weight, so the tension in the string is equal to the weight of block C.
Therefore, the tension on block C is 10.0 kg * 9.8 m/s² = 98 N.
3. The acceleration of block B can be determined by considering the net forces acting on it.
The net force is equal to the tension in the string minus the frictional force.
The frictional force can be calculated as the coefficient of kinetic friction (0.100) multiplied by the normal force.
The normal force is equal to the weight of block B minus the tension in the string.
Using Newton's second law (F = ma), we can solve for the acceleration.
The magnitude of the acceleration of block B is (68.6 N - 0.100 * (7.00 kg * 9.8 m/s² - 68.6 N)) / 7.00 kg = 1.64 m/s².
The direction of the acceleration is toward the right because the tension in the string is greater than the frictional force.
So therefore the magnitude is 1.64 m/s².