A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during the process is.

Options

A. constant and equal to `mg` in magnitude
B. constant and greater than `mg` in magnitude
C. variable but always greater than `mg`.
D. at first greater than `mg`, and later becomes equal to `mg`.

Explanation

To solve the problem of the force of reaction of the ground on a man who is squatting and then stands up, we can analyze the situation step by step.

### Step 1: Understand the Initial Position
When the man is squatting, he is exerting a force on the ground equal to his weight (mg, where m is mass and g is acceleration due to gravity). The ground exerts an equal and opposite force (normal reaction, N) back on him.

**Hint:** Remember Newton's Third Law: for every action, there is an equal and opposite reaction.

### Step 2: Analyze the Transition to Standing
As the man begins to stand up, he needs to exert a force greater than his weight to accelerate upwards. This means that the normal reaction force (N) from the ground must be greater than his weight (mg) during the initial phase of standing.

**Hint:** Consider the forces acting on the man when he is accelerating upwards.

### Step 3: Apply Newton's Second Law
Using Newton's Second Law, we can express the forces when the man is transitioning from squatting to standing:
- The net force acting on the man is given by \( N - mg = ma \), where \( a \) is the upward acceleration.
- Rearranging gives us \( N = mg + ma \). Here, \( N \) must be greater than \( mg \) because \( a \) is positive (the man is accelerating upwards).

**Hint:** Identify that the net force must be greater than zero for the man to accelerate upwards.

### Step 4: Analyze the Final Position
Once the man stands up completely, he is in a state of rest again. At this point, the forces acting on him are balanced:
- The normal force (N) from the ground is equal to his weight (mg), so \( N = mg \).

**Hint:** When at rest, the forces acting on an object are balanced.

### Conclusion
In summary, during the process of standing up, the force of reaction (normal force) from the ground on the man is greater than his weight while he is accelerating upwards and equals his weight when he is standing still. Therefore, we can conclude:

- While squatting and beginning to stand: \( N > mg \)
- While standing still: \( N = mg \)

Thus, the correct option is that the force of reaction of the ground on the man during the process is greater than his weight when he is standing up.

**Final Answer:** The force of reaction of the ground on the man during the process is greater than his weight (N > mg) when he is standing up.