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Question Number 69045 by jagannath19 last updated on 18/Sep/19

Answered by Rio Michael last updated on 18/Sep/19

$$F_{net}=100cos53°$$$$ma\approx 60$$$$2\left(a\right)=60$$$$a=30m/s^2$$

Answered by mr W last updated on 18/Sep/19

$$F_x=100\cos 53°\approx 60N$$$$F_y=100\sin 53°\approx 80N$$$${ma}_x=F_x$$$$\Rightarrow a_x=\frac{F_x}{m}=\frac{60}{2}=30m/s^2$$$${ma}_y=F_y-mg$$$$\Rightarrow a_y=\frac{F_y}{m}-g=\frac{80}{2}-10=30m/s^2$$$$a=\sqrt{a_x^2+a_y^2}=\sqrt{{\left(30\right)}^2+{\left(30\right)}^2}$$$$=30\sqrt{2}$$$$=42.4m/s^2$$

Commented by necxxx last updated on 19/Sep/19

$$wow.Thankyousomuch$$

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