a student attaches a mass to the end of a 0.80 m string. The system will be whirled in a horizontal circular path at 31.5 m/s. The maximum tension the string can withstand without breaking is 250 N. What is the maximum mass the student can use?

To find the maximum mass the student can use without breaking the string, we need to determine the tension in the string when the system is whirling at a speed of 31.5 m/s.

First, we use the formula for centripetal force:

F = (m * v^2) / r

Where:
F is the centripetal force
m is the mass of the object
v is the velocity
r is the radius of the circular path

In this case, the maximum tension in the string must equal the centripetal force to prevent it from breaking. Therefore:

Maximum tension = (m * v^2) / r

Rearranging the formula, we can solve for m:

m = (Maximum tension * r) / v^2

Substituting the given values:

Maximum tension = 250 N
Radius, r = 0.8 m
Velocity, v = 31.5 m/s

m = (250 N * 0.8 m) / (31.5 m/s)^2

Calculating the value:

m = (200 N * 0.8 m) / (993.75 m^2/s^2)
m = 160 N / 993.75 m^2/s^2
m ≈ 0.161 kg

Therefore, the maximum mass the student can use without breaking the string is approximately 0.161 kg.

tension=mass*v^2/r

solve formass.