A 0.40-kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 2.0 m on a frictionless horizontal surface.

If the cord will break when the tension in it exceeds 75 N, what is the maximum speed the ball can have?

May someone please walk me through this problem step by step

ma = T

ma = mv²/R
mv²/R= T
v=sqrt(R•T/m)

Sure! Let's go through this problem step by step:

Step 1: Identify the relevant equation(s):

In order to solve this problem, we need to use the equation for centripetal force:

F = m * a_c

Where:
- F is the centripetal force
- m is the mass of the object
- a_c is the centripetal acceleration

We also know that the centripetal acceleration can be expressed as:

a_c = v^2 / r

Where:
- v is the velocity of the object
- r is the radius of the circular path

Step 2: Determine the maximum tension in the cord:

Since the cord will break when the tension in it exceeds 75 N, we can set the centripetal force (F) equal to 75 N:

F = Tension = 75 N

Step 3: Calculate the centripetal force:

Substituting the given values into the equation F = m * a_c, we have:

Tension = m * (v^2 / r)

Step 4: Substitute the given values into the equation:

Given:
- m (mass of the ball) = 0.40 kg
- r (radius of the circular path) = 2.0 m
- Tension = 75 N

75 N = 0.40 kg * (v^2 / 2.0 m)

Step 5: Solve for v (velocity):

To solve for v, rearrange the equation and isolate v:

v^2 = (75 N * 2.0 m) / 0.40 kg
v^2 = 375 N*m / 0.40 kg
v^2 = 937.5 N * m / kg

Taking the square root of both sides:

v = sqrt(937.5 N * m / kg)

Step 6: Calculate the maximum speed:

Substitute the value of v into the equation:

v = sqrt(937.5 N * m / kg)
v = sqrt(937.5 N * 0.40 kg / kg)
v = sqrt(937.5 N * 0.40)
v ≈ sqrt(375.0)
v ≈ 19.36 m/s

Therefore, the maximum speed the ball can have is approximately 19.36 m/s.

Certainly! To find the maximum speed the ball can have before the cord breaks, we need to examine the tension in the cord at that point.

Here are the steps to solve the problem:

Step 1: Identify the relevant physics principles:
We are dealing with circular motion and tension in a cord. For circular motion, we use the centripetal force equation: Fc = mv^2 / r, where Fc is the centripetal force, m is the mass of the ball, v is the velocity, and r is the radius. In this case, the centripetal force is provided by the tension in the cord.

Step 2: Determine the maximum tension that the cord can withstand:
Given in the problem, the cord can break when the tension exceeds 75 N.

Step 3: Express the relationship between tension and velocity:
We can equate the maximum tension in the cord to the centripetal force: Tension = mv^2 / r.

Step 4: Solve for the maximum velocity:
Rearrange the equation to solve for velocity: v = √(Tr / m), where Tr is the maximum tension the cord can withstand.

Step 5: Plug in the values and calculate:
Now we substitute the given values into the equation to find the maximum speed the ball can have. In this case, the mass is 0.40 kg, the radius is 2.0 m, and the maximum tension is 75 N.

v = √(75 N * 2.0 m / 0.40 kg)

v = √(150 N * 5 m / 0.40 kg)

v = √(750 N*m / 0.40 kg)

v = √(1875 m^2/s^2 / kg)

Therefore, the maximum speed the ball can have is approximately 30.84 m/s.

Please note that the value may vary slightly due to rounding.