A stunt driver gives her car 3600 N to take a jump that has an angle of inclination of 31°. If she and the car together weigh 4700 N, what is the acceleration of the car? (Assume no friction.)
mg=4700Nmg=4700Nmg=4700N
⇒m=47009.8=479.59\Rightarrow m=\dfrac{4700}{9.8}=479.59⇒m=9.84700=479.59 kg
Balancing the forces along the plane,
ma=F−4700sin31οma=F-4700sin31^{\omicron}ma=F−4700sin31ο
⇒a=3600479.59−4700479.59sin31ο\Rightarrow a=\dfrac{3600}{479.59}-\dfrac{4700}{479.59}sin31^{\omicron}⇒a=479.593600−479.594700sin31ο
⇒a=2.554m/s2\Rightarrow a=2.554m/s^2⇒a=2.554m/s2
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