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Under a force of 10i -3j+6k Newton a body of mass 5 kg is displaced from the position 6i+5j-3k to the position 10i-2j+7k. What is the work done?
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John Opella, B.S. Mechanical Engineering, Texas A&M University (2020)
Answered Jul 25
Work equals the dot product of force and displacement. W=F∙dW=F•d
Force is given as 10i−3j+6k10i−3j+6k
Displacement is calculated from the final position minus the initial position.
d=(10i−2j+7k)−(6i+5j−3k)=4i−7j+10kd=(10i−2j+7k)−(6i+5j−3k)=4i−7j+10k
W=(10i−3j+6k)∙(4i−7j+10k)W=(10i−3j+6k)•(4i−7j+10k)
W=(10∗4)+((−3)∗(−7))+(6∗10)=40+21+60=121W=(10∗4)+((−3)∗(−7))+(6∗10)=40+21+60=121
Assuming positions given were in meters, work is 121 Joules.✌✌
HOPE IT HELPS U❤❤❤✌✌
Under a force of 10i -3j+6k Newton a body of mass 5 kg is displaced from the position 6i+5j-3k to the position 10i-2j+7k. What is the work done?
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John Opella, B.S. Mechanical Engineering, Texas A&M University (2020)
Answered Jul 25
Work equals the dot product of force and displacement. W=F∙dW=F•d
Force is given as 10i−3j+6k10i−3j+6k
Displacement is calculated from the final position minus the initial position.
d=(10i−2j+7k)−(6i+5j−3k)=4i−7j+10kd=(10i−2j+7k)−(6i+5j−3k)=4i−7j+10k
W=(10i−3j+6k)∙(4i−7j+10k)W=(10i−3j+6k)•(4i−7j+10k)
W=(10∗4)+((−3)∗(−7))+(6∗10)=40+21+60=121W=(10∗4)+((−3)∗(−7))+(6∗10)=40+21+60=121
Assuming positions given were in meters, work is 121 Joules.✌✌
HOPE IT HELPS U❤❤❤✌✌
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Hope u know addong of scalar quantity
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