K.E=2mv2
also
W = F.d Where F is Force and d is distance
Given that
F= 4000 N this frictional force
m = 2100 Kg
θ= 20.0°
V=3.8 m/s V is the car’s speed at the bottom of the driveway
W=Δ K.E
W=0.5×2100×3.82=15162J
Since the x component of the gravity is
Fx=mgsinФ
so
Fx=2100×9.8sin(20.0°)=7038.77N
And the Net force is
Fnet=Fx−Ff
Fnet=7038.77–4000=3038.77N
And the length of the driveway =FnetW=3038.7715162=4.98m
So the length of the driveway=4.98M
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