Question #241851

A metal car starts from rest and slides 12 m down a chute that is inclined at 40° to the horizontal. It then continues to slide along a horizontal length of the chute and eventually comes to rest. If μ = 0,2 between the chute and the car, calculate the horizontal length of the chute. Use conservation of energy to solve this question.


Clearly show the following values in their respective sections

Section A:

Ek=?

Ep=?

ETR=?

ETE=?

Energy equation=??

v=??


Section B:

Ek=?

Ep=?

ETR=?

ETE=?

Distance=??




1
Expert's answer
2021-09-27T01:44:29-0400

given data

massoftruck=mt=8000kgmass of truck = m_{t} = 8000 kg

speedoftruck=Vt=4m/sspeed of truck =V_{t} = 4 m/s

massofcar=mc=2000kgmass of car = m _{c} = 2000 kg

speedofcar=Vc=16m/sspeed of car = V _{c} = 16 m / s

so 

Vc=16cos30°i+(16sin30°)jV_{c} = -16cos 30\degree i + ( -16 sin 30\degree) j



Vt=4iV_{t} = 4i



- total momentum before collision 

p=mtvt+mcvc=800(4i)+2000(16(0.866)i8j)p = m_{t} v_{t}+ m _{c} v_{c} = 800 (4 i) + 2000(-16(0.866)i-8j)

p=32000i27712i16000jp = 32000 i - 27712 i - 16 000j

p=4288i16000jp = 4288 i - 16000j

p=16564mkg/s|p| = 16564 m kg/s



2-


as after collision they move together & as momentum is conserve

p=pf=16564=(mt+mc)vf|p| = |p_{f}| = 16564 = (m _{t} + m_{c}) v _{f}

vf=p/m=0.04288i1.6jv_{f} = p/m = 0.04288i - 1.6 j

KE1+KE2=mtvt2/2+mcvc2/2=EiKE _{1} + KE _{2} = m_{t} v_{t}^2/2 + m_{c} v_{c}^2 /2 = E_{i}

Ei=4000(4)2+1000(16)2E_{i} = 4000 (4)^2 + 1000(16)^2

Ei=320000JE_{i} = 320000 J

final energy = (mt+mc/2)vf2(m_{t} + m_{c} /2) v_{f}^2

ve2=5000(1.66)2v_{e}^2 =5000(1.66)^2


Ef=13778JE_{f}=13778 J


as EiE_i is not equal to EfE_f thus not elastic collision


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