a, Wt= mgh= 0,2.10.10=20(J)

vt=\(\sqrt{2gh}\)= \(\sqrt{2.10.10}\)=10\(\sqrt{2}\)(m/s)

Wđ= 1/2.m\(v^2\)= 1/2.0,2.\(\left(10\sqrt{2}\right)^2\)=20 (J)

=> Wđ chạm đất =Wt

b, ta có Wđ=Wt=20J

W=Wđ+Wt => W=2Wt 

=> 20 =2mgh' 

=>20= 2.0,2.10.h'

=>h'=5(m)

W= Wđ+ Wt = 1/2m\(v^2\)+ mgh= 1,5mgh+mgh => 37,5= 1,5m. 3.10+m.10.3   => m=0,5

W= Wđ+ Wt = 1/2m\(v^2\)+ mgh  => 37,5= 1/2.0,5.\(v^2\)+ 0,5. 10.3 => v=\(3\sqrt{10}\)

a, Đổi 2 tấn= 2000kg;    v=21,6km/h=6m/s

\(_{v_t=v_o+at}\)=>\(\dfrac{v-v_o}{t}\)=\(\dfrac{6-0}{15}\)= 0,4

 \(\)F=ma => F= 2000.0,4=800(N)

\(v^2-v_o^2=2as\) => \(\dfrac{6^2-0^2}{2.0,4}\)=45 (m)

A=F.s.cos\(\alpha\)= 800.45.cos\(0^o\)= 36000 (J)

P=\(\dfrac{A}{t}\)= \(\dfrac{36000}{15}\)=2400 (J)

b, ADĐL 2 newton: \(\overrightarrow{P}+\overrightarrow{N}+\overrightarrow{F_k}-\overrightarrow{F_{ms}}=m\overrightarrow{a}\)(1)

Chiếu (1) lên như hình vẽ:F-Fms=ma => F-(0,05. 2000.10)=2000.0,4 => F= 1800(N)

A=F.s.cosa= 1800.45=81000 (J)

P=A/t = 81000/15 =5400 (J)

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