Question

A puck of mass m = 0.075 kg is moving in a circle on a horizontal frictionless surface. It is held in its path by a massless string of length l = 0.36 m. The puck makes one revolution every t = 0.65 s. What is the centripetal force acting on the puck?

1) 0.115 N
2) 0.231 N
3) 0.346 N
4) 0.462 N

Answer 1

The centripetal force acting on the puck is 0.462 N.

Step-by-step explanation:

The centripetal force acting on the puck can be calculated using the formula Fc = (m * v^2) / r, where Fc is the centripetal force, m is the mass of the puck, v is the linear velocity of the puck, and r is the radius of the circular path.

In this case, the mass of the puck is 0.075 kg, the linear velocity is calculated by dividing the circumference of the circle (2 * pi * l) by the time it takes to complete one revolution (t), and the radius of the circular path is given as 0.36 m.

Substituting these values into the formula, we have Fc = (0.075 kg * ((2 * pi * 0.36 m) / 0.65 s)^2) / 0.36 m.

Simplifying the equation, we get Fc = 0.462 N. Therefore, the centripetal force acting on the puck is 0.462 N.