Two boxes of fruit on a frictionless horizontal surface are connected by a light string as in the figures below, where m1 = 15 kg and m2 = 17 kg. A force of 58 N is applied to the 17 kg box.
Repeat the problem for the case where the coefficient of kinetic friction between each box and the surface is 0.09.
a) find the acceleration
b) find the tension in the string
To solve this problem, we will use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration (F = m * a). We will also take into account the force of friction and tension in the string.
a) To find the acceleration of the system, we first need to determine the net force acting on it. In this case, the only horizontal force acting on the system is the applied force of 58 N.
1. For the case of no friction:
Since there is no friction, the net force is equal to the applied force:
Net force = 58 N
Next, we need to calculate the total mass of the system:
Total mass = m1 + m2
= 15 kg + 17 kg
= 32 kg
Now, we can use Newton's second law to find the acceleration:
Net force = mass * acceleration
58 N = 32 kg * acceleration
Dividing both sides of the equation by 32 gives us:
acceleration = 58 N / 32 kg
Calculating the value:
acceleration = 1.8125 m/s²
Therefore, the acceleration of the system in the absence of friction is 1.8125 m/s².
2. For the case of a coefficient of kinetic friction of 0.09:
Now, we need to consider the force of friction. The force of friction can be calculated as the product of the coefficient of kinetic friction (μ) and the normal force (N). Since the boxes are connected by a string, the normal force acting on each box is equal to its weight.
Weight of m1 = m1 * g
= 15 kg * 9.8 m/s²
= 147 N
Weight of m2 = m2 * g
= 17 kg * 9.8 m/s²
= 166.6 N
The force of friction on each box is:
Force of friction = μ * Normal force
For m1:
Force of friction on m1 = μ * Weight of m1
= 0.09 * 147 N
= 13.23 N
For m2:
Force of friction on m2 = μ * Weight of m2
= 0.09 * 166.6 N
= 14.994 N
The net force acting on the system is now the difference between the applied force and the force of friction on m2:
Net force = Applied force - Force of friction on m2
= 58 N - 14.994 N
= 43.006 N
Applying Newton's second law, we have:
Net force = mass * acceleration
43.006 N = 32 kg * acceleration
Dividing both sides of the equation by 32 gives:
acceleration = 43.006 N / 32 kg
Calculating the value:
acceleration = 1.3439375 m/s²
Therefore, the acceleration of the system with a coefficient of kinetic friction of 0.09 is 1.3439375 m/s².
b) To find the tension in the string, we can consider the forces acting on m1 and m2 separately.
For m1:
The tension in the string is equal to the force of friction on m1:
Tension = Force of friction on m1
= 13.23 N
For m2:
The tension in the string is equal to the force of friction on m2 plus the mass of m2 times its acceleration:
Tension = Force of friction on m2 + (m2 * acceleration)
= 14.994 N + (17 kg * 1.3439375 m/s²)
= 14.994 N + 22.85578125 N
= 37.84978125 N
Therefore, the tension in the string is 37.84978125 N with a coefficient of kinetic friction of 0.09.