25.4k views
3 votes
A puck of mass m=0.085 kg is moving in a circle on a horizontal frictionless surface. It is held in its path by a massless string of length l=0.24 m. The puck makes one revolution every t=0.45 s. What is the centripetal force acting on the puck?

1) 0.085 N
2) 0.19 N
3) 0.38 N
4) 0.85 N

User Elmonty
by
8.1k points

1 Answer

3 votes

Final answer:

The centripetal force acting on the puck is approximately 0.38 N.

Step-by-step explanation:

The centripetal force acting on the puck can be calculated using the formula:

Fc = (m * v2) / r

Where:
- Fc is the centripetal force
- m is the mass of the puck
- v is the linear velocity of the puck
- r is the radius of the circular path

Given:
- m = 0.085 kg
- v = 2πr / t (using the formula for linear velocity in terms of the radius and time to complete one revolution)
- r = 0.24 m
- t = 0.45 s

Substituting the given values into the formula:

Fc = (m * ((2πr / t)2)) / r

Calculating:

Fc = (0.085 * ((2π*0.24 / 0.45)2)) / 0.24

Fc ≈ 0.38 N

Hence, the centripetal force acting on the puck is approximately 0.38 N.

User Jack Zach Tibbles
by
7.8k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories