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(b) $m_{suitcase}=20.0kg,F=35.0N$

$∑F_{x}=ma_{x}:−20.0N+Fcosθ=0$

$∑F_{y}=ma_{y}:+n+Fsinθ−F_{g}=0$

$Fcosθ=20.0N$

$cosθ=35.0N20.0N =0.571$

$θ=55.2_{0}$

(c) With $F_{g}=(20.0kg)(9.80m/s_{2})$,

$n=F_{g}−Fsinθ=[196N−(35.0N)(0.821)]$

$n=167N$

Solve any question of Laws of Motion with:-

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