a string 0.5 m long is used to whirl a 1 kg stone in a horiziontal circle at a uniform velocity of 5m/s what is the tension on the string?
the tension in the string provides the force for the centripetal acceleration of the stone
f = m * v^2 / r = 1 * 5^2 / .5
the force is in Newtons
To find the tension in the string, we can use the centripetal force formula. The centripetal force is the force that keeps an object moving in a circular path. In this case, it is the tension in the string that provides the centripetal force to keep the stone moving in a horizontal circle.
The centripetal force is given by the formula: F = (m * v^2) / r
Where:
F is the centripetal force
m is the mass of the stone (1 kg)
v is the velocity of the stone (5 m/s)
r is the radius of the circular path (0.5 m)
Plugging in the values, we get:
F = (1 kg * (5 m/s)^2) / 0.5 m
First, let's calculate 5^2 which is equal to 25:
F = (1 kg * 25 m^2/s^2) / 0.5 m
Now, let's calculate (1 kg * 25 m^2/s^2), which is equal to 25 kg*m^2/s^2:
F = (25 kg*m^2/s^2) / 0.5 m
Finally, let's divide 25 kg*m^2/s^2 by 0.5 m:
F ≈ 50 N
Therefore, the tension on the string is approximately 50 Newtons.