A 2150-kg truck is traveling along a straight, level road at a constant speed of 55.0 km/h when the driver removes his foot from the accelerator. After 21.0 s, the truck's speed is 33.0 km/h. What is the magnitude of the average net force acting on the truck during the 21.0 s interval?
A) 2250 N
B) 626 N
C) 1890 N
D) 972 N
E) 229 N
Guide On Rating System
Vote
To find the average net force acting on the truck, we first need to convert the speeds from km/h to m/s:
Initial speed: 55.0 km/h = 55.0 * (1000 m/3600 s) = 15.28 m/s
Final speed: 33.0 km/h = 33.0 * (1000 m/3600 s) = 9.17 m/s
Next, we calculate the acceleration of the truck:
a = (vf - vi) / t
a = (9.17 m/s - 15.28 m/s) / 21 s
a = -0.35 m/s^2
Now we can calculate the net force acting on the truck:
F = ma
F = 2150 kg * (-0.35 m/s^2)
F = -750 N
The negative sign indicates that the force is in the opposite direction to the motion of the truck. The magnitude of the average net force is 750 N.
Therefore, the closest answer choice is:
B) 626 N
Got a question with my answer?
Cheapmath.com is completely free!
Asking question is free.
Getting answers is free.
+99