Adamaged 1200-kg car is being towed by a truck. neglecting the friction, air drag, and rolling resistance, determine the extra power required (a) for constant velocity on a level road, (b) for constant velocity of 50 km/h on a 30° (from horizontal) uphill road, and (c) to accelerate on a level road from stop to 90 km/h in 12 s.

(a) 0 W

In this situation, the car is moving on a level road at constant velocity. This means that its acceleration is zero:

a = 0

But this also means that the net force acting on the car is zero, according to Newton's second law:

F = ma = 0

The power required to move the car is

P = Fv

where F is the force applied to the car and v the velocity. Since there are no air drag/frictional forces, the force applied to the car is exactly zero: so, the extra power needed to keep the car in motion at constant velocity is also zero.

(b) 81815 W

In this situation, the car is moving on an uphill road, so there is a component of the weight parallel to the incline acting agains the direction of motion of the car. This component of the weight is given by

where

m = 1200 kg is the mass of the car

g = 9.81 m/s^2 is the acceleration of gravity

is the angle of the incline

Substituting,

So the truck must exerts an additional force of F = 5886 N to keep the car in motion at constant velocity.

The velocity of the car is

v = 50 km/h = 13.9 m/s

So the extra power needed is

P = Fv = (5886 N)(13.9 m/s) = 81,815 W

(c) 31250 W

Here the car accelerates to a final velocity of

v = 90 km/h = 25 m/s

from an initial velocity of

u = 0

So the work done on the car is equal to its change in kinetic energy, so:

The extra power erogated by the truck therefore must be

where

t = 12 s

is the time taken

Substituting,

kinetic

i'm so sorry if this is wrong i'm only in 6th grade xd

i just did an experiment like this in my science class. also the answer is yes, it will get separated because they are two opposites meaning if they are opposite they can not attract.

hope this .!