A puck (mass m_1 = 1.35 kg) slides on a frictionless table as shown in the figur

Question

A puck (mass m_1 = 1.35 kg) slides on a frictionless table as shown in the figure below. The puck is tied to a string that runs through a hole in the table and is attached to a mass m_2 = 4.1 kg. The mass m_2 is initially at a height of h = 4.1 m above the floor with the puck traveling in a circle of radius r = 1.08 m with a speed of 4.1 m/s. The force of gravity then causes mass m_2 to move downward a distance 0.41 m. What is the new speed of the puck? What is the change in the kinetic energy of the puck?

Explanation / Answer

given

m1 = 1.35 kg

r1 = 1.08 m

v1 = 4.1 m/s

r2 = 1.08 - 0.41

= 0.67 m

v2 = ?

a) Apply conservation of angular momentum

m1*v1*r1 = m1*v2*r2

==> v2 - v1*r1/r2

= 4.1*1.08/0.67

= 6.61 m/s

b) change in kinetic energy of the puck = 0.5*m1*(v2^2 - v1^2)

= 0.5*1.35*(6.61^2 - 4.1^2)

= 18.1 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.