A stone with a mass of 10 kg sits on the ground. Gravity acts on the stone at a rate of 9.8 m/s2. What is the normal force acting on the stone, keeping it at rest?(1 point)

Question

Responses

9.8 N
9.8 N

10 N
10 N

98 N
98 N

1.02 N

Answer 1

To determine the normal force acting on the stone, we can use the formula for weight (force due to gravity), which is given by:

\[ F = m \cdot g \]

where:

• \( F \) is the force,
• \( m \) is the mass of the object, and
• \( g \) is the acceleration due to gravity (9.8 m/s²).

For the stone:

• \( m = 10 , \text{kg} \)
• \( g = 9.8 , \text{m/s²} \)

Now, substituting the values:

\[ F = 10 , \text{kg} \times 9.8 , \text{m/s²} = 98 , \text{N} \]

Since the stone is at rest and there are no other vertical forces acting on it (assuming it is on a flat surface), the normal force (the force exerted by the ground to support the stone) must equal the weight of the stone. Therefore, the normal force is:

98 N.

So the correct response is:

98 N.