Answer:
The force that the 10 kg block applies to the 5 kg block is **20 N**. The correct option is **c)**.
To explain why, we can use Newton's third law of motion, which states that for every action, there is an equal and opposite reaction¹². That means the forces of action and reaction are always equal and opposite, and they act on two different objects³.
In this case, the 30 N force is the action force that acts on the 10 kg block. The reaction force is the force that the 10 kg block exerts on the 5 kg block. According to Newton's third law, these two forces must be equal and opposite, so the force that the 10 kg block exerts on the 5 kg block is also 30 N, but in the opposite direction.
However, this is not the net force on the 5 kg block, because there is another force acting on it: the weight of the 10 kg block. The weight of the 10 kg block is given by multiplying its mass by the acceleration due to gravity, which is approximately 9.8 m/s^2. Therefore, the weight of the 10 kg block is:
$$W = mg = (10 \text{ kg}) (9.8 \text{ m/s}^2) = 98 \text{ N}$$
This weight acts downward on the 5 kg block, so it reduces the net force on it. The net force on the 5 kg block is given by subtracting the weight of the 10 kg block from the force exerted by it:
$$F_{\text{net}} = F - W = (30 \text{ N}) - (98 \text{ N}) = -68 \text{ N}$$
The negative sign indicates that the net force is downward. To find the magnitude of this force, we take its absolute value:
$$|F_{\text{net}}| = |-68 \text{ N}| = 68 \text{ N}$$
This is the net force on the 5 kg block due to both the 10 kg block and gravity. However, this is not what we are looking for. We want to find only the force that the 10 kg block applies to the 5 kg block, without gravity. To do this, we need to add back the weight of the 10 kg block to the net force:
$$F = F_{\text{net}} + W = (-68 \text{ N}) + (98 \text{ N}) = 30 \text{ N}$$
This is the same as the force that we found earlier using Newton's third law. However, this is still not our final answer, because this is the force that acts on both blocks as a system. We need to find only the force that acts on one block: the 5 kg block.
To do this, we need to use another concept from Newton's second law of motion, which states that the net force on an object is equal to its mass times its acceleration⁴. That means we can find the acceleration of both blocks by dividing their net force by their mass:
$$a = \frac{F_{\text{net}}}{m}$$
Since both blocks are connected and move together, they have the same acceleration. Therefore, we can use either block to find their common acceleration:
$$a = \frac{F_{\text{net}}}{m} = \frac{-68 \text{ N}}{(10 \text{ kg}) + (5 \text{ kg})} = -4.53 \text{ m/s}^2$$
The negative sign indicates that their acceleration is downward. To find only the force that acts on one block: the 5 kg block, we need to multiply its mass by its acceleration:
$$F = ma = (5 \text{ kg}) (-4.53 \text{ m/s}^2) = -22.65 \text{ N}$$
The negative sign indicates that this force is also downward. To find its magnitude, we take its absolute value:
$$|F| = |-22.65 \text{ N}| = 22.65 \text{ N}$$
This is almost our final answer, but we need to round it to one significant figure, since that is how many significant figures are given in the options. To do this, we look at the second digit after the decimal point and see if it is greater than or equal to five. If it is, we round up; if not, we round down. In this case, it is five, so we round up:
$$|F| \approx 23 \text{ N}$$
This is our final answer, but none of the options match it exactly. Therefore, we need to choose the closest option to our answer, which is **c)** 20 N. This is the force that the 10 kg block applies to the 5 kg block.