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)
Responses
9.8 N
9.8 N
98 N
98 N
1.02 N
1.02 N
10 N
The normal force acting on the stone to keep it at rest is equal in magnitude and opposite in direction to the gravitational force acting on it. According to Newton's second law, F = m * a, where F is force, m is mass, and a is acceleration. In this case, the acceleration is due to gravity, which is 9.8 m/s².
Using the formula F = m * a, we can calculate the gravitational force: F = 10 kg * 9.8 m/s² = 98 N.
Therefore, the normal force acting on the stone, which is equal in magnitude and opposite in direction to the gravitational force, is 98 N.
So the answer is:
98 N
The normal force acting on an object at rest on a flat surface is equal in magnitude and opposite in direction to the force of gravity acting on the object.
In this case, the force of gravity acting on the stone is given as 9.8 m/s². Therefore, the normal force acting on the stone, keeping it at rest, is also 9.8 N.
So the correct option is: 9.8 N
To find the normal force acting on the stone, we need to use Newton's second law, which states that the net force acting on an object is equal to the product of its mass and acceleration.
Since the stone is at rest on the ground, we know that the acceleration is zero. Therefore, the net force acting on the stone must also be zero.
The force of gravity acting on the stone is given by the formula:
force = mass * acceleration due to gravity
Substituting the given values into the formula:
force = 10 kg * 9.8 m/s^2
force = 98 N
Since the net force is zero, the normal force must be equal in magnitude but opposite in direction to the force of gravity.
Therefore, the normal force acting on the stone is 98 N.