Answer :
Answer:
This is their velocity on the top of the slope [tex]v=3.7 \frac{m}{s}[/tex]
Explanation:
The problem doesn't have final question however is an example of conservation of energy
[tex]Kf+(Ug)f=Ki+(Ug)j+W[/tex]
And given from the problem to know the final velocity
[tex]Yi=0\\vi=5.3 \frac{m}{s}\\ m=25kg\\F=8.0N\\d=100m\\Yf=4m[/tex]
[tex]Kf+(Ug)f=Ki+(Ug)j+W[/tex]
[tex]\frac{1}{2}*m*vf ^{2}+m*g*Yf=\frac{1}{2}*m*vi^{2}+m*g*Yi+W\\\frac{1}{2}*m*vf ^{2}+m*g*Yf=\frac{1}{2}*m*vi^{2}+W\\\frac{1}{2}*m*vf ^{2}=\frac{1}{2}*m*vi^{2}+W-m*g*Yf\\vf ^{2}=vi^{2}+2*\frac{W}{m}+2*g*Yf\\vf ^{2}=(5.3\frac{m}{s})^{2} +\frac{2*F*d}{m}-2*9.8\frac{m}{s^{2}}*4m\\vf ^{2}=(5.3\frac{m}{s})^{2} +\frac{2*8.0N*100m}{25kg}-2*9.8\frac{m}{s^{2}}*4m\\vf=\sqrt{13.69 \frac{m^{2} }{s^{2} } } \\vf=3.7 \frac{m}{s}[/tex]