ch09-p033 - 33. (a) By energy conservation, the speed of...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
26.6 m/s 7.0 m/s 19.6 m/s, vv v ′′ =− = = and the magnitude of the impulse becomes 3 | | | | (90kg)(19.6m/s) 1.76 10 N s. Jp m v m v =Δ = Δ = = × (d) The corresponding average force would be 3 5 avg 3 1.76 10 N s 3.52 10 N. 5.0 10 s J F t ×⋅ == × Δ× 33. (a) By energy conservation, the speed of the passenger when the elevator hits the floor is 22 1 2 2(9.8 m/s )(36 m) 26.6 m/s. 2 mv mgh v gh = ¡ = Thus, the magnitude of the impulse is 3 | | | | (90 kg)(26.6 m/s)
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online