Answer:
Step-by-step explanation:
We can use conservation of energy to solve this problem. At the lowest point of the vertical circle, all the potential energy is converted into kinetic energy. Therefore, we can find the speed of the mass at this point:
Potential energy at highest point = Kinetic energy at lowest point
mgh = (1/2)mv^2
where m = 30.0 kg, g = 9.81 m/s^2 (acceleration due to gravity), h = 20.0 m (twice the radius of the circle), and v is the speed of the mass at the lowest point.
Solving for v, we get:
v = sqrt(2gh) = sqrt(2 x 9.81 x 20.0) = 19.8 m/s
Since the mass moves at a constant speed of 5.0 m/s, it will travel a distance of:
distance = speed x time = 5.0 m/s x 5.0 s = 25.0 m
However, we need to find the horizontal distance traveled by the mass. At the lowest point, the weight of the mass provides the centripetal force required for circular motion. When the mass breaks off, it will move in a straight line tangent to the circle at the point of detachment. Therefore, we can find the horizontal distance traveled by the mass using the equation:
distance = vt
where v is the horizontal component of the velocity and t is the time taken.
The horizontal component of the velocity is given by:
v_horizontal = v x cos(theta)
where theta is the angle between the velocity vector and the horizontal axis. At the lowest point, the velocity vector is vertical, so theta = 90 degrees. Therefore, we have:
v_horizontal = v x cos(90) = 0
This means that the mass will not move horizontally from the lowest point, and the distance traveled in the horizontal direction is zero.