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# A ball of mass m is attached to a light string of length l. ******* PHYSICS?

A ball of mass m is attached to a light string of length l. The ball is held with the string horizontal as shown in the diagram below. Then, the ball is released. What is the tension in the string when the ball is at its lowest point?

A. zero

B. mg

C. 2mg

D. 3mg

E. 4mg

http://www.webassign.net/poe1/2005-mc-02.gif

DAMNT IT THIS QUESTION PISSES ME OFF IDK WHY I JUST DONT GET IT

ISNT IT T - MG = MA?

T= M(G + A)

TENSION IS POSITIVE IN THE CENTER AND WEIGHT IS GOING DOWN

### 2 Answers

- JánošíkLv 78 years agoFavorite Answer
At its release point, the ball is a distance L higher than its lowest point.

The potential energy at its point of release is:

Ep = m × g × L

At its lowest point, that energy in converted to kinetic energy, so the velocity of the ball is:

m × v² / 2 = m × g × L

v² / 2 = g × L

v² = 2 × g × L

v = √(2 × g × L)

The centripetal acceleration at that point is

Ac = v² / L

Ac = (2 × g × L) / L

Ac = 2 × g

The total acceleration at that point is the centripetal acceleration plus the acceleration by gravity, so

a = 2×g + g

a = 3×g

and the tension in the rope is

T = m × a

T = m × 3×g

T = 3mg

- Anonymous8 years ago
F=ma

2)Then specify each force: F t -F g =ma 3)move around: F t =ma+F g (or you can make in mg at this point)

4)Simplify:F t =m(a+g)

As for the answer, idk, without numbers for mass or length.