# Assume that three blocks portrayed below move on a frictionless surface and that a 42 N force acts on the 3.0 kg block. Determine (a) the acecleration given the system, (b) the tension inthe cord connecting the 3.0kg and the 1.0 kg blocks, and (c) the force exerted by the 1.0kg block on the 2.0kg block.

The 1 kg box has the 2 kg box right next to it on the right side and the cord is attached between the 1kg box and the 3 kg box which is over to the right.

So far I got the acceleration to be 7.0 m/s^2 but I can't get a tension. How do I go about getting it?

13 years ago

## a.7m/s

b.21n
c.14n toward the light

8 months ago

## To determine the tension in the cord connecting the 3.0 kg and the 1.0 kg blocks, we can consider the forces acting on the system.

Let's denote the acceleration of the system as "a" and the tension in the cord as "T".

(a) Acceleration of the system:
Given that a 42 N force acts on the 3.0 kg block:
Using Newton's second law of motion, we can calculate the acceleration of the system.
F = ma
42 N = (3.0 kg + 1.0 kg) * a
42 N = 4.0 kg * a
a = 42 N / 4.0 kg
a = 10.5 m/s^2

So, the acceleration of the system is 10.5 m/s^2.

(b) Tension in the cord:
To find the tension in the cord, we need to consider the forces acting on the 3.0 kg and 1.0 kg blocks.

For the 3.0 kg block:
The only force acting on it is the tension in the cord.
Using Newton's second law in the vertical direction for the 3.0 kg block, we can express the equation as:
T - (3.0 kg * 9.8 m/s^2) = (3.0 kg * a)
T - 29.4 N = (3.0 kg * 10.5 m/s^2)
T - 29.4 N = 31.5 N
T = 31.5 N + 29.4 N
T = 60.9 N

So, the tension in the cord is 60.9 N.

(c) Force exerted by the 1.0 kg block on the 2.0 kg block:
To determine the force exerted by the 1.0 kg block on the 2.0 kg block, we can consider Newton's third law of motion, which states that the forces of action and reaction are equal in magnitude and opposite in direction.

The force exerted by the 1.0 kg block on the 2.0 kg block is equal in magnitude but opposite in direction to the force exerted by the 2.0 kg block on the 1.0 kg block.

Therefore, the force exerted by the 1.0 kg block on the 2.0 kg block is 42 N in the opposite direction.

So, the force exerted by the 1.0 kg block on the 2.0 kg block is 42 N.

8 months ago

## To find the tension in the cord connecting the 3.0 kg and the 1.0 kg blocks, you can follow these steps:

1. Start by analyzing the forces acting on each individual block and the system as a whole.

2. For the 3.0 kg block: We know that a force of 42 N is acting on it. Since the surface is frictionless, the only force acting on it is the tension in the cord.

3. For the 1.0 kg block: The tension in the cord is acting on it, and it exerts a force on the 2.0 kg block.

4. For the 2.0 kg block: Along with the force exerted by the 1.0 kg block, there is no other external force acting on this block.

5. Determine the acceleration of the system. In this case, you mentioned that the acceleration of the system is 7.0 m/s^2.

Now, to find the tension in the cord:

Step 1: Calculate the net force on the 3.0 kg block.
The net force can be obtained using the formula:
Net force = mass x acceleration

For the 3.0 kg block:
Net force = 3.0 kg x 7.0 m/s^2
Net force = 21 N

Since the only force acting on the 3.0 kg block is the tension in the cord, the tension is equal to the net force acting on it.

Step 2: Find the tension in the cord by considering the 1.0 kg block.
The tension in the cord also acts on the 1.0 kg block. Therefore, we can equate the tension to the force exerted by the 1.0 kg block.

Tension = Force exerted by the 1.0 kg block on the 2.0 kg block

Since the 1.0 kg block exerts an equal and opposite force on the 2.0 kg block, the tension in the cord is equal to the force exerted by the 1.0 kg block on the 2.0 kg block.

Therefore, the tension in the cord is 21 N.

To summarize:
(a) The acceleration of the system is 7.0 m/s^2.
(b) The tension in the cord connecting the 3.0 kg and the 1.0 kg blocks is 21 N.
(c) The force exerted by the 1.0 kg block on the 2.0 kg block is also 21 N (based on Newton's Third Law of Motion).

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