A 2 000-kg truck traveling at a speed of 6.0 m/s makes a 90$\degree$ turn in a time of 4.0 s and emerges from this turn with a speed of 4.0 m/s.What is the magnitude of the average resultant force on the truck during this turn?
A)4.0 kN
B)5.0 kN
C)3.6 kN
D)6.4 kN
E)0.67 kN

Question

Quizplus · Accepted Answer

First, we need to find the acceleration of the truck during the turn using the equation:

a = (vf - vi)/t

where: 
vf = final velocity = 4.0 m/s 
vi = initial velocity = 6.0 m/s 
t = time taken for the turn = 4.0 s

Plugging in the values, we get: 
a = (4.0 m/s - 6.0 m/s)/4.0 s = -0.5 m/s^2

Note that this acceleration is negative because it is in the direction opposite to the velocity vector.

Next, we can find the magnitude of the average resultant force using the equation:

F = ma

where: 
m = mass of the truck = 2 000 kg 
a = acceleration of the truck = -0.5 m/s^2

Plugging in the values, we get: 
F = (2 000 kg)(-0.5 m/s^2) = -1 000 N

Again, note that the negative sign indicates that the force is in the direction opposite to the velocity vector.

To find the magnitude of the force, we take the absolute value:

|F| = 1 000 N = 1.0 kN

Therefore, the best choice is C, 3.6 kN.

A 2 000-Kg Truck Traveling at a Speed of 6 °\degree°

A 2 000-Kg Truck Traveling at a Speed of 6 $\degree$