A package, mass M=2kg is dropped vertically from a height h=0.5m onto a horizont

Question

A package, mass M=2kg is dropped vertically from a height h=0.5m onto a horizontal conveyor belt which is moving at a uniform speed v= 2m/s.

Assume the box experiences no vertical bounce during impact, which last t=0.5s.

coefficient of kinetic friction between the box and belt=0.4

A) what is the average vertical force to which the box is subjected during impact?

B) after impact box is being accelerated by the conveyor belt taking a distance d=1m.

What force is responsible for this change in motion?

C) calculate the thermal engery between belt and box?

iF SOMEONE COULD EXPLAIN AND BE THOROUGH THAT'D BE AMAZING

Explanation / Answer

a)

avg force is change in momentum/time

here velocity of block just before impact is v=sqrt(2*g*h)=sqrt(2*9.8*.5)=3.1305 m/s

so momentum intial =2*v=6.261 kg m/s

force=6.26099/.5=12.52198 N

b) here frictional force is responsible for accelaration.

c)frictional force is uN=.4*mg=.4*2*9.8=7.84N

thermal energy =work done against friction=F*D=

7.84*1=7.84 joules

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