Ryan E. answered 10/05/21
Professional Engineer with MS in ECE and CISSP
FP=8.40N
mA=10.5kg
mB=7.00kg
mT=mA+mB=10.5kg+7.00kg=17.50kg
FP=mTa
a=?
a=FP/mT=8.40N/17.5kg=0.48m/s2
a=0.48m/s2
James D.
asked 10/05/21Alex is asked to move two boxes of books in contact with each other and resting on a frictionless floor. He decides to move them at the same time by pushing on box A with a horizontal pushing force FP = (+8.40) N . Here A has a mass mA = 10.5 kg and B has a mass mB = 7.00 kg. The contact force between the two boxes is FC and N denotes the normal force.
a) What is the magnitude of the acceleration of the two boxes? = ? m/s^2
Ryan E. answered 10/05/21
Professional Engineer with MS in ECE and CISSP
FP=8.40N
mA=10.5kg
mB=7.00kg
mT=mA+mB=10.5kg+7.00kg=17.50kg
FP=mTa
a=?
a=FP/mT=8.40N/17.5kg=0.48m/s2
a=0.48m/s2
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