Two boxes are connected by a cord running over a pulley as shown in the picture below. Box I of mass 8.0 kg is on the top of the table; the coefficient of kinetic friction between box I and the table is 0.10. Box II has a mass of 15.0 kg. What is the tension in the cord?

35 N

57 N

69 N

93 N

150 N

57N

57

To find the tension in the cord, we first need to calculate the net force acting on each box.

For Box I:

1. Determine the weight of Box I: Fg1 = m1 * g, where m1 is the mass of Box I and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Fg1 = 8.0 kg * 9.8 m/s^2 = 78.4 N

2. Determine the friction force acting on Box I: Ff1 = μ * Fg1, where μ is the coefficient of kinetic friction.
Ff1 = 0.10 * 78.4 N = 7.84 N (opposite the direction of motion)

3. Determine the net force on Box I: Fnet1 = Fg1 - Ff1
Fnet1 = 78.4 N - 7.84 N = 70.56 N

For Box II:

1. Determine the weight of Box II: Fg2 = m2 * g, where m2 is the mass of Box II.
Fg2 = 15.0 kg * 9.8 m/s^2 = 147 N

2. Determine the net force on Box II: Fnet2 = Fg2
Fnet2 = 147 N

Now, since Box I and Box II are connected by a cord passing over a pulley, they experience equal magnitudes of tension. Therefore, the tension in the cord is equal to the net force on either box.

Tension = Fnet1 = Fnet2 = 70.56 N.

Therefore, the tension in the cord is 70.56 N.

To find the tension in the cord, we need to consider the forces acting on the two boxes.

First, let's consider Box I. The weight of Box I is given by the equation: W = mg, where m is the mass of Box I (8.0 kg) and g is the acceleration due to gravity (9.8 m/s^2). So the weight of Box I is: W = (8.0 kg)(9.8 m/s^2) = 78.4 N.

Next, let's consider the forces acting on Box I. There are two forces acting on Box I: the weight (78.4 N) and the friction force opposing the motion. The friction force is given by the equation: F_friction = coefficient of kinetic friction × normal force. The normal force is given by the equation: F_normal = mass × gravity = (8.0 kg)(9.8 m/s^2) = 78.4 N. So the friction force is: F_friction = (0.10)(78.4 N) = 7.84 N.

The net force acting on Box I is given by the equation: F_net = F_applied - F_friction, where F_applied is the applied force. In this case, the net force is zero since the box is not accelerating. So we have: 0 = F_applied - 7.84 N. Solving for F_applied, we find: F_applied = 7.84 N.

Since Box I and Box II are connected by a cord running over a pulley, they experience the same tension force. So the tension in the cord is also 7.84 N.