A string under a tension of 43.0 N is used to whirl a rock in a horizontal circle of radius 2.45 m at a speed of 20.9 m/s on a frictionless surface as shown in the figure below. As the string is pulled in, the speed of the rock increases. When the string is 1.00 m long and the speed of the rock is 49.0 m/s, the string breaks. What is the breaking strength, in newtons, of the string?
tension=mv^2/r initial. figure mass.
now, final tension
tensionbreaking= m v^2/r you know mass, v, and r.
To find the breaking strength of the string, we need to calculate the maximum tension force applied on the string when it breaks.
Given:
Tension in the string when the rock is whirling in a circle with a radius of 2.45 m at a speed of 20.9 m/s is 43.0 N.
The length of the string when it breaks is 1.00 m.
The speed of the rock when the string breaks is 49.0 m/s.
When an object moves in a circle at a constant speed, it experiences a centripetal force directed towards the center of the circle. This force is provided by the tension in the string.
The centripetal force can be calculated using the equation:
F = (mv²) / r
Where:
F = Centripetal force
m = Mass of the object
v = Velocity of the object
r = Radius of the circular path
In this case, we don't know the mass of the rock, but we can safely ignore it because it does not affect the tension force.
So, rearranging the equation, we can calculate the tension force:
T = m * v² / r
Now, let's calculate the tension force when the rock is whirling at 20.9 m/s.
T₁ = m * v₁² / r
T₁ = m * (20.9 m/s)² / (2.45 m)
T₁ = m * 435.61 m²/s² / 2.45 m
T₁ = m * 177.63 N
Now, let's calculate the tension force when the rock is whirling at 49.0 m/s (at the moment the string breaks).
T₂ = m * v₂² / r
T₂ = m * (49.0 m/s)² / (1.00 m)
T₂ = m * 2401 m²/s² / 1.00 m
T₂ = m * 2401 N
The breaking strength of the string is equal to the maximum tension force exerted on it before it breaks. Therefore, at T₂, the tension force is at its maximum.
Therefore, the breaking strength of the string is 2401 N.