A stone with a mass of 1 kg is attached to one end of a string 121 cm long. The stringwill break if its tension just exceeds 400 N. The stone is whirled in a horizontal circle ona frictionless table top. The other end of the string is kept fixed. Find the maximumspeed in m/s of the stone, it can attain without breaking the string.
Question
A stone with a mass of 1 kg is attached to one end of a string 121 cm long. The stringwill break if its tension just exceeds 400 N. The stone is whirled in a horizontal circle ona frictionless table top. The other end of the string is kept fixed. Find the maximumspeed in m/s of the stone, it can attain without breaking the string.
Solution
The tension in the string is provided by the centripetal force, which is given by the formula:
F = m*v^2/r
where: F is the force (in newtons), m is the mass (in kilograms), v is the velocity (in meters per second), and r is the radius (in meters).
We know that the string will break if the tension exceeds 400 N, so we can set F equal to 400 N. We also know that the mass of the stone is 1 kg, and the length of the string (which is the radius of the circle) is 121 cm, or 1.21 m. Substituting these values into the formula, we get:
400 N = 1 kg * v^2 / 1.21 m
Solving for v, we get:
v^2 = 400 N * 1.21 m / 1 kg v^2 = 484 m^2/s^2
Taking the square root of both sides, we find that the maximum speed of the stone is:
v = sqrt(484 m^2/s^2) = 22 m/s
So, the stone can attain a maximum speed of 22 m/s without breaking the string.
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