26.6 m/s 7.0 m/s 19.6 m/s,vvv′′′=− =−=and the magnitude of the impulse becomes 3||||(90kg)(19.6m/s) 1.76 10 N s.Jpmvmv=Δ =Δ ==≈×⋅(d) The corresponding average force would be 35avg31.76 10 N s3.52 10 N.5.0 10 sJFt−×⋅==≈×Δ×33. (a) By energy conservation, the speed of the passenger when the elevator hits the floor is22122(9.8 m/s )(36 m)26.6 m/s.2mvmghvgh=¡=Thus, the magnitude of the impulse is 3||||(90 kg)(26.6 m/s)
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