Question
Question: A horizontal force of \(500\text{ N}\) pulls two masses \(20kg\) and \(10kg\)(lying on a frictionles...
A horizontal force of 500 N pulls two masses 20kg and 10kg(lying on a frictionless table) connected by a light string. What is the tension in the string?
A. 200 NB. 300 NC. 400 ND. 3500N
Solution
Hint: Whenever there is an external force acting on a system, the system will be accelerated. If the two blocks are connected by a string, then the tension in the string will be directed away from both the blocks. If we take the forces acting on the blocks separately we will be able to solve for the tension.
Complete step by step answer:
When a force is applied to a system, it will be accelerated. Since the two masses are tied together by a string, the string will experience a tension that is directed away from both the masses.
The force acting on the 10kg mass is the tension T that is acting away from the block and the acceleration of the system due to the applied force. So we can write,
T=ma
T=10a ……equation (1)
Coming to the block of mass of 20kg, the forces acting on the block are, the tension T which acts away from the block, the force of 500 N acting opposite to the tension and finally the acceleration of the mass along the direction of force applied. So, we can write,
F−T=ma
From equation (1), we have T=10a, so the above equation can be written as,
500 N−10a=20a⇒500 N=(10+20)a
∴a=350 m/s2
Substituting the value of a in equation (1), we get
T=3500N
So the tension in the string is T=3500N.
So the answer to the question is option (D).
Note:
When the system is at rest or moving with constant velocity, it means that the external force acting on the system is zero. If friction was present, the frictional force would try to reduce the acceleration of the system.
Tension is always directed away from the body. We can see tension as a pull, which pulls the bodies it is attached to. The tension disappears when the string slackens.