A clay stone with mass m = 5 kg travelling to the East with speed 10 m/s. It hit

Question

A clay stone with mass m = 5 kg travelling to the East with speed 10 m/s. It hits an initially stationary wooden block with mass M = 20 kg. The clay sticks to the wooden block and they move as one on the rough floor. The wooden block plus clay travel a distance S = 5m before they stop. Find: a) What is the speed of block plus clay after the collision? b) How much energy was lost during the collision? c) What is the coefficient of friction between the floor and the wooden block? d) How much energy what dissipated as heat by friction?

Explanation / Answer

(a)conserving momentum

5*10=(20+5)*v

we got v=2 m/s

(b) initial energy=0.5*5*100=250 J

final energy=0.5*25*4=50 J

Energy lost=200 J

(c)let the coefficient of friction be u

friction force=u*mg=u*25*9.8

work done by friction=change in kinetic energy

u*25*9.8*5=50

u=0.04

(d)Energy dissipated as heat=50 J

we got

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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