If a person pushes on a block and it travels a distance of 6.0m before stopping. How far would the block travel if it were pushed with the same force, but on the moon(which has 1/6 of the earth's gravity).
To figure out how far the block would travel on the moon with 1/6 of Earth's gravity, we first need to understand the relationship between force, mass, and acceleration.
Newton's second law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (F = m * a). In this case, the force applied is the same, but the acceleration will be different due to the change in gravity.
1. First, we need to determine the acceleration on the moon. Since the moon's gravity is 1/6 of Earth's gravity, the acceleration on the moon will also be 1/6. Therefore, the acceleration on the moon is (1/6 * 9.8 m/s^2) = 1.63 m/s^2.
2. Next, we can use the equation for distance traveled to calculate the distance the block would travel on the moon. The equation is:
distance = (initial velocity * time) + (1/2 * acceleration * time^2)
In this case, the initial velocity of the block is 0 m/s because it starts from rest.
3. Since the block travels the same distance on both Earth and the moon (6.0 m), we can use the same time for both calculations.
distance = (0 * t) + (1/2 * 1.63 * t^2)
4. Solving this equation will give us the distance the block would travel on the moon. Rearranging the equation to solve for distance yields:
distance = (1/2 * 1.63 * t^2)
Plugging in the known values:
6.0 m = (1/2 * 1.63 * t^2)
5. Solving for t (time) will give us the time it takes for the block to travel the given distance on the moon.
t^2 = (6.0 m) / (0.815 m/s^2)
t^2 = 7.36
t = sqrt(7.36)
t ≈ 2.71 seconds
6. Now, we can use the time calculated in step 5 to find the distance the block would travel on the moon.
distance = (0 * 2.71 s) + (1/2 * 1.63 m/s^2 * (2.71 s)^2)
distance ≈ 1.69 meters
Therefore, if the block were pushed with the same force on the moon, it would travel approximately 1.69 meters before stopping.