A block of mass M 1 = 3.5 kg moves with velocity v 1 = 6.3 m/s on a frictionless

Question

A block of mass M1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. It collides with block of mass M2 = 1.7 kg which is initially stationary. The blocks stick together and encounter a rough surface. The blocks eventually come to a stop after traveling a distance d = 1.85 m . What is the coefficient of kinetic friction on the rough surface?

?k =

A block of mass M1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. It collides with block of mass M2 = 1.7 kg which is initially stationary. The blocks stick together and encounter a rough surface. The blocks eventually come to a stop after traveling a distance d = 1.85 m . What is the coefficient of kinetic friction on the rough surface? ?k =

Explanation / Answer

M1*V1 = (M1+M2) * V

V =M1*V1 /(M1+M2) =3.5*6.3/(3.5+1.7) = 4.24 m/s

on rough surface

0.5*M*V^2 - u *M*g*s = 0

u = 0.5*M*V^2 / Mg = 0.5*V^2/g*s = 0.5*4.24^2 / 9.8*1.85 = 0.496

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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