Answer to Question #273534 in Mechanics | Relativity for Lili

Question #273534
  1. A 2.1 × 103 kg car starts from rest at the top of a driveway that is sloped at an angle of 20.0° with the horizontal. An average friction force of 4.0 × 103 N impedes the car’s motion so that the car’s speed at the bot- tom of the driveway is 3.8 m/s. What is the length of the driveway?
1
Expert's answer
2021-11-30T18:15:51-0500

K.E=mv22K.E=\frac{mv^2}{2}


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=15162JW = 0.5×2100×3.8^2=15162 J

Since the x component of the gravity is

Fx=mgsinФFx = mg sinФ

so

Fx=2100×9.8sin(20.0°)=7038.77NFx = 2100×9.8sin(20.0° ) =7038.77 N

And the Net force is

 Fnet=FxFfF_{net}=F_x-F_f

Fnet=7038.774000=3038.77NF_{net}= 7038.77 – 4000 = 3038.77 N

And the length of the driveway =WFnet=151623038.77=4.98m\frac{W}{F_{net}} =\frac{15162}{3038.77} = 4.98 m

So the length of the driveway=4.98M


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