Vật lý 10 - Tính CÔng

By Trang Ánh Nam -

 

 

Giải:

Bài 10: chọn chiều dương hướng lên ta có 2 chuyển động.

I.                  Chuyển động lên nhanh dần đều:

A=Ak+AP  ( Ak là công của Trục, AP là công của trọng lực)

óFs.cosα=AK+P.s.cosβ

óm.a.scosα= AK+mgcosβ

ó AK = m.a.scosα – m.gs.cosβ

óAK=100.0,8.10.1 -100.10.10.(-1) (do vật đi lên nên α=0 độ và β=180 độ)

óAK=800+10000=10800

II.               Chuyển động lên chặm dần đều:

Vận tốc của vật khi bắt đầu chuyển động chậm.

v2-vo2=2as

=>v2=2.0,8.10=16

=>v=4 m/s

Gia tốc của vật khi chuyển động chậm dần:

Do sau đó vật dừng lại nên v’=0

Quảng đường vật đi được khi chuyển động chặm dần điều là:

Áp dụng công thức công:

A=A’k+AP

ó A’K = m.a.scosα – m.gs.cosβ

óA’K=100.(-0,4).20.1 – 100.10.20.(-1)

óA’K=19200

Vậy tổng công là: ATrục=AK+A’K=10800+19200=30000 J                 

                                                                                                             

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