A stone with a mass of 0.80 kgkg is attached to one end of a string 0.90 mm long. The string will break if its tension exceeds 50.0 NN. The stone is whirled in a horizontal circle on a frictionless tabletop; the other end of the string remains fixed. (a) Draw a free-body diagram of the stone.
(b) Find the maximum speed the stone can attain without the string breaking.

(a) the image is attached below

(b) v = 0.237 m/s

Explanation:

(a) The free-body diagram can be observed in the attached image

(b) To find the maximum speed you take into account the centripetal force:

m: mass = 0.80kg

v: tangential speed of the stone = ?

r: radius of the circular trajectory of the stone = 0.90mm = 0.90*10^-3 m

The force cannot exceed 50.0N, then you have:

hence, the maximum speed the stone can attain without breaking the string is 0.237m/s

a displacement distance is size of displ

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s