Question

Two boxes of fruit on a frictionless horizontal surface are connected by a light string as in Figure P4.85, where m1 = 10 kg and m2 = 20 kg. A force of 50 N is applied to the 20-kg box. (a) Determine the acceleration of each box and the tension in the string. (b) Repeat the problem for the case where the coefficient of kinetic friction between each box and the surface is 0.10.

Answer 1

To determine the acceleration of each box and the tension in the string, analyze the forces acting on the system. In the first case (no friction), the acceleration of each box will be the same, and the tension in the string will be equal to the applied force. In the second case (with friction), the acceleration and tension will be lower due to the frictional forces.

Step-by-step explanation:

To determine the acceleration of each box and the tension in the string, we can analyze the forces acting on the system. Since the surface is frictionless, the only forces acting on the system are tension and the applied force. By applying Newton's second law, we can write the following equations:

For box 1 (m1): m1 × a1 = T - F

For box 2 (m2): m2 × a2 = F - T

Solving these equations simultaneously, we can find the values of a1, a2, and T. In the first case (no friction), the acceleration of each box will be the same, and the tension in the string will be equal to the applied force. In the second case (with friction), we need to consider the frictional force on each box when determining the acceleration and tension in the string. The acceleration and tension will be lower due to the frictional forces. To account for the friction, we can modify the equations as follows:

For box 1 (m1): m1 × a1 = T - F - f1

For box 2 (m2): m2 × a2 = F - T - f2

Where f1 and f2 are the frictional forces acting on each box. Using the coefficient of kinetic friction (μmk), we can calculate the frictional forces as:

f1 = μmk × m1 × g

f2 = μmk × m2 × g

Substituting the values of f1 and f2 into the equations and solving them simultaneously will give us the acceleration and tension in the string.

Answer 2

The acceleration of each box is 2.5 m/s² and the tension in the string is 25 N. When there is a coefficient of kinetic friction, the acceleration is (50 N - (0.10 * 20 kg * 9.8 m/s²)) / 30 kg and the tension is 10 kg * acceleration.

Step-by-step explanation:

To determine the acceleration of each box and the tension in the string, we can use Newton's second law and the concept of tension. Since the boxes are connected, they will have the same acceleration. The force acting on the 20 kg box is 50 N, and the mass of the 10 kg box is 10 kg. We can use the equation F = ma to find the acceleration of the 20 kg box: 50 N = 20 kg * a. Solving for a, we get a = 2.5 m/s².

Since the boxes have the same acceleration, the acceleration of the 10 kg box is also 2.5 m/s². Now, to find the tension in the string, we can use the equation F = ma again, but this time for the 10 kg box. The only force acting on the 10 kg box is the tension, so we have T = 10 kg * 2.5 m/s². Solving for T, we get T = 25 N.

(b) When there is a coefficient of kinetic friction between each box and the surface, we need to take into account the frictional force. The frictional force can be calculated as the product of the coefficient of kinetic friction and the normal force. The normal force is equal to the weight of the box, which is m2 * g. In this case, m2 is 20 kg and g is 9.8 m/s². The frictional force is then 0.10 * (20 kg * 9.8 m/s²). We can subtract this force from the force applied to the 20 kg box to find the net force: 50 N - (0.10 * 20 kg * 9.8 m/s²). This net force is equal to the mass of the system (m1 + m2) times the acceleration. So we have (m1 + m2) * a = 50 N - (0.10 * 20 kg * 9.8 m/s²), where m1 + m2 is 30 kg. Solving for a, we get a = (50 N - (0.10 * 20 kg * 9.8 m/s²)) / 30 kg. Now we can use this acceleration to find the tension in the string, using the equation F = ma. The force acting on the 10 kg box is the tension, so we have T = 10 kg * a.