Posted by **kim** on Tuesday, November 11, 2008 at 10:06pm.

1. A cart of weight 20 N is accelerated across a level surface at .15 m/s^2. What net force acts on the wagon? (g=9.8m/s^2)

2. An automobile of mass 2000 kg moving at 30 m/s is braked suddenly with a constant braking force of 10000 N. How far does the car travel before stopping?

3. An airplane of mass 1.2 X10^4 kg tows a glider of mass .6X10^4 kg. The airplane propellers provide a net foward thrust of 3.6X10^4 N. What is the gliders acceleration?

how do i solve these?

- physics -
**drwls**, Wednesday, November 12, 2008 at 1:37am
You solve #1 and #3 by applying Newton's Second Law, F = m a.

For the second problem, you first need to solve for the required deceleration rate. a = F/m = 5 m/s^2

The time it takes to stop is

t = Vo/a = (30 m/s)/5 m/s^2 = 6 s

The distance travelled while stopping is that stopping time multiplied by the average velocity (Vo/2),

15 m/s * 6 s = 90 m

- physics -
**Aigerim**, Saturday, December 13, 2014 at 8:47am
From Newtons' 2nd Law of Motion,

F = ma

where

F = braking force = 10,000 N

m = mass of the car = 2,000 kg.

a = acceleration

Substituting values,

-10,000 = 2,000(a)

NOTE the negative sign attached to the braking force. This simply denotes that the direction of this particular force is opposite that of the motion of the automobile.

Solving for "a",

a = -10,000/2,000

a = -5 m/sec^2

The negative value of the acceleration means that when the brakes were applied, the car, obviously, was slowing down until it came to a complete stop.

The next formula to use is

Vf^2 - Vo^2 = 2as

where

Vf = final velocity = 0 (when the car finally stops)

Vo = initial velocity = 30 m/sec (given)

a = acceleration = -5 m/sec^2 (as calculated above)

s = stopping distance of car

Substituting values,

0 - 30^2 = 2(-5)(s)

-900 = -10s

and solving for "s",

s = 900/10

s = 90 meters

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