As part a of the drawing shows, two blocks are connected by a rope that passes over a set of pulleys. One block has a weight of m1 = 562 N, and the other has a weight of m2 = 758 N. The rope and the pulleys are massless and there is no friction.

(a) What is the acceleration of the lighter block?

(b) Suppose that the heavier block is removed, and a downward force of 758 N is provided by someone pulling on the rope, as part b of the drawing shows. Find the acceleration of the remaining block.

(c) Explain why the answers in (a) and (b) are different.

a) Well, it seems we have a classic case of "heavy versus light" here. The acceleration of the lighter block can be determined by applying Newton's second law (F = ma). Since there is no friction and the rope and pulleys are massless, the tension in the rope will be the same throughout. So, let's call the tension T.

For block m1, we have the equation T - m1g = m1a, where g is the acceleration due to gravity. Rearranging the equation, we get T = m1g + m1a.

For block m2, we have the equation m2g - T = m2a. Substituting the value of T from the previous equation, we get m2g - m1g - m1a = m2a.

Now, let's solve for a. Rearranging the equation, we get (m2 - m1)g = (m1 + m2)a. Dividing both sides by (m1 + m2), we get a = (m2 - m1)g / (m1 + m2).

(b) Okay, now let's remove the heavier block and imagine someone pulling downward on the rope with a force of 758 N. In this case, the lighter block is the only one left, so we can use the equation T - m1g = m1a, where T is the tension in the rope.

Since the force applied (758 N) is greater than the weight of the block (562 N), we can conclude that the tension is equal to the applied force: T = 758 N.

Applying this to the equation, we get 758 N - m1g = m1a. Rearranging, we find a = (758 N - m1g) / m1.

(c) The answers in (a) and (b) are different because in (a), both blocks are connected and have to move together. The acceleration of the lighter block is influenced by the weight of the heavier block. However, in (b), only the lighter block is present, so its acceleration is affected directly by the applied force. Thus, the removal of the heavier block changes the acceleration of the remaining block. It's like going from a two-person dance party to a solo dance-off – the moves are going to be different!

(a) To find the acceleration of the lighter block, we need to consider the forces acting on it. Since the rope and pulleys are massless and there is no friction, the tension in the rope is the same throughout. Let's call this tension "T".

The weight of the lighter block is m1 = 562 N. According to Newton's second law, the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the net force acting on the lighter block is the tension in the rope (T) minus its weight (m1g, where g is the acceleration due to gravity).

Net force = T - m1g

We also know that the net force equals the mass of the object (m1) multiplied by its acceleration (a). Therefore, we can write:

T - m1g = m1a

Rearranging this equation, we have:

a = (T - m1g) / m1

Substituting the values m1 = 562 N and g = 9.8 m/s^2 (acceleration due to gravity) into the equation, we need the value of tension T, which is equal to the weight of the heavier block. Since the weight of the heavier block is m2 = 758 N, we have:

a = (758 N - 562 N x 9.8 m/s^2) / 562 N

Now, we can calculate the acceleration of the lighter block.

(b) In this case, when the heavier block is removed, the downward force of 758 N is provided by someone pulling on the rope. Since the rope and pulleys are still massless and there is no friction, the tension in the rope is equal to the force applied (758 N) throughout the system.

So now, the net force acting on the remaining block is equal to the tension in the rope (T) minus its weight (m1g). We can consider it as:

Net force = T - m1g

Since the net force is also equal to the mass of the object (m1) multiplied by its acceleration (a), we can write:

T - m1g = m1a

Rearranging this equation, we have:

a = (T - m1g) / m1

Substituting the values m1 = 562 N and g = 9.8 m/s^2 into the equation, we need the value of tension T, which is equal to the downward force applied (758 N). So, we have:

a = (758 N - 562 N x 9.8 m/s^2) / 562 N

Now, we can calculate the acceleration of the remaining block.

(c) The reason why the answers in (a) and (b) are different is because of the weight of the heavier block. In part (a), the weight of the heavier block contributes to the net force acting on the system, which affects the acceleration of the lighter block. However, in part (b), when the heavier block is removed and a downward force is applied externally, the weight of the heavier block no longer contributes to the net force. The tension in the rope is now equal to the external force applied, and the resulting acceleration of the remaining block is different.