All Physics geniuses or good A2 students help me with this problem please!?

Question

A 0.5kg stone on the end of a length of a string is whirled round in a horizontal circle, radius 2m on a level frictionless table. The stone moves at a constant speed of 7m/s.Calculate

(a) The stones acceleration, stating its direction,

(b) the tension in the string,

(c)the work done by the tension during ten revolutions, explaining your answer

I've attempted part a and b using a=v2/r and tension = mg, i dont know if im right in these 2 but i dont have any idea how to do (c).
Any help is appreciated.

Answer 1

(a)
- firstly, identify from the question -> circular motion, therefore the acceleration is towards the centre.
- secondly, we are provided with data, which is:
v = 7ms^-1 - [linear velocity]
r = 2m - [radius]
a = ( v^2 ) / r
a = (7)^2 / 2 = 24.5ms^-2 - [acceleration]
(b)
- now over here, two things come into the picture; firstly, the question says that the motion is on a level frictionless table, therefore the reaction force balances out the weight component (mg).
secondly, if we ignore the reaction component; then:
- let the angle the string makes with the normal to the motion (vertical) be 'x'
now T being the tension in the string can be resolved into 2 vectors: Tcos(x) and T sin(x)
now, Tcos(x) = m g = 0.5 x 9.81 = 4.905 and
Tsin(x) = ( m v^2 ) / r =a = ( m ( v^2 ) / r ) = 0.5 x 24.5 = 12.25
now x can be found out by, dividing Tsin(x) / Tcos(x) = tan(x)
tan(x) = 12.25 / 4.905
(x) = 68.2 degrees
now finding T = 4.905 / cos(x) = 13.2 N
(c)
- again, identifying from the question -> circular motion, therefore workdone = 0. Why?
Because, newton's second law states that, force is directly proportional to acceleration and is always acting in the same direction. now since acceleration is towards centre so is the force (centripetal force - causing the object in circular motion).
workdone = force x displacement (in direction of force) or:
W = F s cos(y)
where y = angle the force makes with horizontal and object, in circular motion 'y' always be in 90 degrees, making cos(y) = 0, therefore workdone 0, as force is always acting perpendicular to the motion and does no work in the direction of motion of object.

Done!! |-)...

Answer 2

You are right with (a). Only the direction of the stone's velocity is changing. The acceleration is directed towards the centre of the circular path and is called centripetal acceleration having magnitude v^2/r.
You are wrong with (b). The acceleration due to gravity is vertical and have nothing to do with the horizontal motion on a frictionless table. The tension in the string provides the necessary centripetal force.
(c). The displacement of the stone is along the tangent at the point concerned and the tension is along the radial line. That is displacement is always perpendicular to the tension. The work done is therefore zero.
Therefore,T = mv^2/r = 0.5 x 7^2 / 2 = 12.25 N

Answer 3

a) v^2/R
b) m v^2/R
c) zero. F and s are at 90 degrees.

That will be £5 please.