Question

求帮忙做几道高中物理题谢谢。

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Answer 1

1(a). Vmax=at=(F-f)*t/m=(F-μmg)*t/m=[220-(0.1*72*9.8)]*8/72=16.6
1(b). According to the conservation of energy, W(F)=W(f)
=> F*S(8s)=f*S(all)
=>S(all)=F*[(1/2)*a*(t^2)]/f=F*(1/2)*[(F-μmg)/m]*(t^2)/μmg=207.08m

2(a) First, we analyze the force on the wood block.
To resolve the force of gravity, let F1 be the force along the ramp and F2 be the force perpendicular to the ramp.
F1=mgsinθ; F2=mgcosθ
Therefore, the net force on the wood block along the ramp is F1+f=F1+μ*F2

The vertical height for the wood block to reach is h=s*sinθ
=> h=[(v0^2)/(2*a)]*sinθ=(v0^2)*sinθ/[2*(F1+μ*F2)/m]=(v0^2)*sinθ/[2*(gsinθ+μgcosθ)]=6.11m
2(b). According to the conservation of energy, Ek0-Ekt=f*(2s)
From the last question, we know
f=μmgcosθ, s=(v0^2)/[2*(gsinθ+μgcosθ)]
Therefore, we have
[1/2m(V0^2)]-[1/2m(Vt^2)]=f*(2s)
=> (V0^2)-(Vt^2)=μgcosθ*4*{(v0^2)/[2*(gsinθ+μgcosθ)]}
Solve the equation and we get Vt=8.41m/s

Answer 2

。。。英语真好，还做英文的，没事干了吧

Answer 3

207.2