No, this is not correct. See my previous post.

[James Prado]

Homework Statement

A truck (1200kg) is traveling south at 22m/s
a) what is the net force required to bring the truck to a stop in 330m?
b) what is the cause of this net force? (im assuming the answer is friction due to brakes)

Mass= 1200kg (1.2*10^4)
Velocity= 22m/s [south]
What force is required to stop the object (1200kg) in 330 meters?

Homework Equations

[/B]
anything needed to solve the equation?

The Attempt at a Solution

ΣF= m*a
=1200kg * a

a= ( 22 m/s [south] ) / Δt

330m/22m= Δt
t=15 seconds

a= 22/15
a = 1.4666667

"a" is now the number of meters per second required to decelerate to reach fnet = 0 ?

Unsure if this is on the right track or extremely off what I need to do for a question like this

 

[rude man]

(a) Think energy conservation. Kinetic energy change to heat.
(b) you're right.

 

[vela]

Homework Statement

A truck (1200kg) is traveling south at 22m/s
a) what is the net force required to bring the truck to a stop in 330m?
b) what is the cause of this net force? (im assuming the answer is friction due to brakes)

Mass= 1200kg (1.2*10^4)
Velocity= 22m/s [south]
What force is required to stop the object (1200kg) in 330 meters?

Homework Equations

[/B]
anything needed to solve the equation?

The Attempt at a Solution

ΣF= m*a
=1200kg * a

a= ( 22 m/s [south] ) / Δt

OK up to here.

330m/22m= Δt
t=15 seconds

$x = vt$ only works when there's no acceleration.

a= 22/15
a = 1.4666667

"a" is now the number of meters per second required to decelerate to reach fnet = 0 ?

Unsure if this is on the right track or extremely off what I need to do for a question like this