Two blocks with masses 5 kg and 2 kg are connected
(via frictionless, massless pulleys) by a light,
inextensible string, as shown. There is no friction
between the 5 kg block and the horizontal surface on
which it moves.
The tension T in the string supporting the 2 kg block is
5 kg
A 7 N B 14 N C 20 N D 28 N
net force: 2g
Acceleration=force/totalmass=2g/7=19.6/7 m/s^2
thension= 2(9.8-2.8)=2*7=14N
To find the tension in the string supporting the 2 kg block, we'll use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration.
Let's denote the tension in the string as T, the mass of the 5 kg block as m1, and the mass of the 2 kg block as m2.
Since the pulleys are frictionless and the string is light and inextensible, the tension throughout the string is constant.
For the 5 kg block:
The weight of the 5 kg block is given by W1 = m1 * g, where g is the acceleration due to gravity (~ 9.8 m/s^2).
Therefore, W1 = 5 kg * 9.8 m/s^2 = 49 N.
Since there is no friction between the 5 kg block and the horizontal surface, the only horizontal force acting on it is the tension, T.
Therefore, the net force acting on the 5 kg block is F1 = T.
Using Newton's second law, we can write the equation as F1 = m1 * a1, where a1 is the acceleration of the 5 kg block.
Therefore, T = m1 * a1.
For the 2 kg block:
The weight of the 2 kg block is given by W2 = m2 * g.
Therefore, W2 = 2 kg * 9.8 m/s^2 = 19.6 N.
Since the 2 kg block is connected to the 5 kg block via the same string, the tension in the string supporting the 2 kg block is the same as the tension in the string supporting the 5 kg block.
Therefore, T = F2, where F2 is the net force acting on the 2 kg block.
Using Newton's second law, we can write the equation as F2 = m2 * a2, where a2 is the acceleration of the 2 kg block.
Therefore, F2 = 2 kg * a2.
Now, since the two blocks are connected by the same string, they will have the same acceleration (a1 = a2 = a), assuming the string is taut and there is no slipping between the string and the pulleys.
So, T = m1 * a and T = F2 = 2 kg * a.
Since the tension T is the same in both cases, we have:
m1 * a = 2 kg * a.
Now, we can cancel out the acceleration from both sides of the equation:
m1 = 5 kg.
Therefore, we have:
5 kg * a = 2 kg * a.
After canceling out the acceleration (a), we get:
5 kg = 2 kg.
This equation is not true since 5 kg does not equal 2 kg.
Therefore, there is an error in the question or the given options for the tension in the string supporting the 2 kg block.
To find the tension T in the string supporting the 2 kg block, we need to consider the forces acting on both blocks and set up the equations of motion.
First, let's consider the 5 kg block. The only force acting on it is its weight (mass × acceleration due to gravity). We'll refer to this force as F1.
F1 = 5 kg × 9.8 m/s² (acceleration due to gravity) = 49 N
Since there is no friction between the 5 kg block and the horizontal surface, there is no horizontal force acting on it.
Next, let's consider the 2 kg block. The tension in the string is denoted by T and acts upwards. The weight of the 2 kg block (F2) is 2 kg × 9.8 m/s² = 19.6 N.
Now, due to the connected string, the tension T is the same on both sides. Therefore, we can write the equation:
T = F1 + F2
Substituting the values we found earlier:
T = 49 N + 19.6 N = 68.6 N
So, the tension T in the string supporting the 2 kg block is 68.6 N.
None of the given answer choices (7 N, 14 N, 20 N, 28 N) is the correct answer.