Physics
posted by christian .
A 75.0kg bobsled is pushed along a horizontal surface by two athletes. After the bobsled is pushed a distance of 4.5m starting from the rest, its speed is 6.0m/s. Find the magnitude of the net force on the bobsled.

a= v²/2s
F=ma 
Fd=0.5mv^2
F(4.5)=0.5(75)(6.0)^2
F=0.5(75)(6.0)^2/4.5
F=300N 
Given:
m=75kg
d=4.5m
v=6.00m/s
angle=180
Required:
F=?
Formrula:
W=change of KE
Same as:
0.5mv^2=Fdcos()
Solve for F
F=0.5mv^2/dcos()
F=0.5(75kg)(6.00m/s)^2/(4.5)(cos 180)
F=225/4.5
F=50N
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