A spring initially has no mass attached (far left in the figure above). When a m

Question

A spring initially has no mass attached (far left in the figure above). When a mass (m1) is attached, the spring is stretched (middle in the figure above) and the mass hangs motionless from the spring.

1. For the mass on the spring (m1) what forces are acting on the mass as it hangs? What is the net force on the mass? (Remember, the mass is at rest.)



2. A similar mass (m2) hangs from a string (far right in the figure above). What forces are acting on the mass as it hangs? What is the net force on the mass? (Hint: imagine the string is stretched just a little, like the spring. Remember the mass is at rest. )

Explanation / Answer

My response is inserted underneath each question. 1. For the mass on the spring (m1) what forces are acting on the mass as it hangs? What is the net force on the mass? Since the mass is motionless, we know acceleration = a = 0. So the net force acting on the mass also must be equal to zero in the equation F = m * a (Total Force = mass * acceleration). The forces acting on the mass are gravity acting downward and the spring tension acting upward. 2. A similar mass (m2) hangs from a string. What forces are acting on the mass as it hangs? What is the net force on the mass? Same answer as question 1 except the magnitude of the individual forces will be different due to the different amount of mass. 3. Now imagine moving the mass on a string a few degrees to one side and holding it still with your hand (So if you let go of the mass, the string would swing back and forth).s What forces are acting on the mass now? Which direction (if any) is the net force? The forces acting on the mass are gravity acting downward and the tension of the string. However, since you have changes in both velocity and direction, acceleration is not equal to zero. So this implies at any given moment the net force on the mass is also not equal to zero. The next force acting on the object will be tangent to the circle made by the radius of the string in the direction the mass is moving (centipetal force). Also note that for one very brief instant at the end of the swing stroke the net forces are zero on the mass since the mass is motionless making acceleration = 0. 4. Now imagine letting go. In the instant after you let go, what forces are acting on the mass? Which direction (if any) is the total force? If this question is a continuation of #3, the answer is the same as #3 since in the after you let go the forces created by you holding the mass are no longer acting on the mass and are removed from the equation. 5. Our second experiment involves timing how long it takes a ball to fall from a certain height. Given the time t it takes a ball to fall a distance y, starting from rest and experiencing only the force of gravity, how can you calculate g, the acceleration due to gravity? The basic equation for position of a constant acceleration is: y - y(0) = v(0)*t +0.5*a*t^2; since your acceleration is gravity substitute g for a y - y(0) = v(0)*t +0.5*g*t^2; V(0) = 0 y - y(0) = 0.5*g*t^2; to simplify the math establish positive y as downward as this will make both g and y - y(0) positive numbers. Also, establish the starting point y(0) = 0. Now solve for g. This gives: 2 * [y-y(0)] / t^2 = g; y(0) = 0 2*y / t^2 = g; y = distance the ball fell and t = time required which per the question are given values. I hope this helps.

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