A 900 kg car is traveling at 20 m/s along the road. What force must be applied to the car to stop it in a distance of 30 m2 Assume a
constant change in velocity.
A. 6000 N
B. 12,000 N
C. 7000 N
D. 18,000 N
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Respuesta :

Answer:

A. 6000 N

Explanation:

v²=u²+2as

0²=20²+2x30xa

-400=60a

a=-400/60

a =-6.667m/s²

f =ma

f = 900 x 6.667 = 6003N

F = 6000N

Answer:

A. 6000 N

Explanation:

Set Up: Let the origin be at the location of the car when the force is applied & the +x direction be in the direction the car is traveling.  

List the known & unknown quantities:  

   mass of the car m = 900 kg

  υx = 0 since the force stops the car  

  υ0x = 20 m/s

  x – x0 = 30 m  

  Fx = force = ?  

Solve: First find the acceleration of the car, ax:  

υx2 = υ0x2 + 2ax(x – x0)

ax= υx2⎯ υ0x2/2(x⎯x0) = (0 m/s)2⎯(20 m/s)2/2(30 m) =⎯6.67 m/s2  

Now find the force using Newton’s second law.  

∑Fx=max=(900 kg) (⎯6.67 m/s2)=⎯6000 N

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