A light rope is attached to a block with a mass of 6 kg that rests on a horizontal, frictionless surface. THe horizontal rope passes over a frictionless, massless pulley, and a block of mass m is suspended from the other end. When the blocks are released, the tension in the rope is 18 N. a) what is the acceleration of the 6 kg block ? b) the mass m of the hanging block?
a) is 3 m/s
b) is 2.65 kg
I don't get the equations I should have used
YOU SHOULD USE THE EQUATION WHICH IS LETTER B.
To solve this problem, we can use Newton's second law of motion and the concept of equilibrium. Here's how you can approach it:
a) Determine the acceleration of the 6 kg block:
- Start by considering the forces acting on the 6 kg block. The only force in the horizontal direction is the tension in the rope, which is 18 N.
- Since there is no friction, the net force acting on the block is equal to mass times acceleration (F = ma).
- Plug in the mass of the block (6 kg) and the tension in the rope (18 N) into the equation: 18 N = 6 kg * a.
- Solve for acceleration (a) by dividing both sides of the equation by 6 kg: a = 18 N / 6 kg = 3 m/s².
- The acceleration of the 6 kg block is 3 m/s².
b) Determine the mass (m) of the hanging block:
- Consider the forces acting on the hanging block. It is being pulled downward by gravity with a force equal to its weight (mg).
- The force pulling upward is the tension in the rope, which is 18 N.
- Since the system is in equilibrium, the net force acting on the hanging block is zero (the blocks are not accelerating up or down).
- Write the equation for equilibrium: 18 N - mg = 0.
- Rearrange the equation to solve for the mass (m): m = 18 N / g, where g is the acceleration due to gravity (approximately 9.8 m/s²).
- Plug in the value of g and simplify: m = 18 N / 9.8 m/s² ≈ 1.84 kg.
Therefore,
a) The acceleration of the 6 kg block is 3 m/s².
b) The mass (m) of the hanging block is approximately 1.84 kg.