One end of a 1.0-m long string is fixed; the other end is attached to a 2.0-kg stone. The stone swings in a vertical circle, passing the bottom point at 4.0 m/s. The tension force of the string at this point is about:

T = 51.6 N

Explanation:

We know that maximum tension in vertical circular motion is at the bottom position and thus;

T − Mg = F_c = Mv²/r

Where F_c is centripetal force

Since T − Mg = Mv²/r

Let's make T the subject;

T = Mg + Mv²/r

Where;

T is tension of the string

M is mass of stone

v is speed of stone

r is length of chord

g is acceleration due to gravity which is 9.8 m/s²

From the question,

M = 2Kg

v = 4 m/s

r = 1m

Plugging in the relevant values we can obtain T;

T = M(g + v²/r) = 2(9.8 + 4²/1)

T = 2(9.8 + 16)

T = 51.6N

an obstruction

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.