A hockey puck that has a mass of 170 g is tied to a light string and spun in a circle of radius 1.35 m (on frictionless ice). The string breaks under a tension that exceeds 5.00 N.

What is the maximum speed of the puck without breaking the string? (Express your answer to three significant figures.)

Well, if the puck wants to avoid breaking the string, it needs to take it easy and not exceed the tension limit. So, let's see how fast it can go without crossing that line.

The tension in the string can be calculated using the centripetal force formula, which is Tension = (Mass x Speed^2) / Radius. In this case, we want that tension to be less than 5.00 N.

Let's plug in the numbers and solve for the speed:

5.00 N = (0.170 kg x Speed^2) / 1.35 m

Now, let's solve for Speed:

Speed^2 = (5.00 N x 1.35 m) / 0.170 kg

Speed^2 = 39.7059 m^2/s^2

Speed ≈ 6.291 m/s

So, the maximum speed the puck can have without breaking the string is approximately 6.291 m/s. Just remember, slow and steady wins the race, or in this case, avoids shattering stringy disasters!

To find the maximum speed of the puck without breaking the string, we can use the concept of centripetal force.

The tension in the string provides the centripetal force required to keep the puck in circular motion. When the tension exceeds 5.00 N, the string will break.

The centripetal force is given by the formula:

F = (mv^2)/r

Where:
F is the centripetal force
m is the mass of the puck (170 g = 0.170 kg)
v is the velocity of the puck (maximum speed)
r is the radius of the circle (1.35 m)

We know that the tension in the string should not exceed 5.00 N, so we can set up the inequality:

Tension in the string ≤ 5.00 N

Substituting F with mv^2/r, we get:

(mv^2)/r ≤ 5.00

Rearranging the equation and solving for v^2, we get:

v^2 ≤ (r * Tension in the string) / m

v^2 ≤ (1.35 m * 5.00 N) / 0.170 kg

v^2 ≤ 39.7059 m^2/s^2

Taking the square root of both sides, we have:

v ≤ √(39.7059) m/s

v ≤ 6.304 m/s

Therefore, the maximum speed of the puck without breaking the string is approximately 6.304 m/s.

To find the maximum speed of the puck without breaking the string, we can start by considering the forces acting on the puck when it is spinning in a circle.

The only force acting on the puck is the tension in the string. This tension provides the centripetal force required to keep the puck moving in a circle. The centripetal force can be found using the equation:

F = m * v^2 / r

Where:
F is the centripetal force,
m is the mass of the puck,
v is the velocity of the puck, and
r is the radius of the circle.

In this case, we need to find the maximum speed of the puck, so we want to find the maximum value of v.

To identify the maximum value of v, we need to consider the maximum tension that the string can support without breaking. The problem statement tells us that the maximum tension is 5.00 N.

Therefore, we can set the tension force equal to the centripetal force:

5.00 N = 0.170 kg * v^2 / 1.35 m

Rearranging the equation to solve for v, we have:

v^2 = (5.00 N * 1.35 m) / 0.170 kg

v^2 = 39.7059 m^2/s^2

Taking the square root of both sides, we find:

v = √(39.7059 m^2/s^2)

v ≈ 6.304 m/s

Therefore, the maximum speed of the puck without breaking the string is approximately 6.304 m/s.

tension=mass*v^2/r

5=.170*v^2/1.35

v= sqrt(5*1.35/.170)=6,30 m/s check that with your calculator