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?

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

To solve these problems, you can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration (F = ma).

1. To find the net force acting on the wagon, use the equation F = ma. The mass of the wagon is not given, so we can use the weight and acceleration due to gravity to find it. Weight (W) is equal to the mass (m) multiplied by the acceleration due to gravity (g). Therefore, the mass (m) of the wagon is given by W/g. Given that the weight is 20 N and g is 9.8 m/s^2, the mass of the wagon is 20/9.8 = 2.04 kg. Now we can calculate the net force using F = ma, where the mass is 2.04 kg and the acceleration is 0.15 m/s^2.

2. To find the stopping distance of the car, we can use the equation F = ma. The force of friction acting on the car while braking is equal to the product of mass and acceleration. Rearranging the equation, we have a = F/m. Given that the mass of the car is 2000 kg and the braking force is 10000 N, we can calculate the acceleration. Then, you can use the equation v^2 = u^2 + 2as, where v is the final velocity (which is 0 m/s as the car stops), u is the initial velocity (30 m/s), a is the acceleration, and s is the stopping distance.

3. To find the glider's acceleration, we can use the concept of Newton's second law, F = ma. The net force acting on the glider is the force of the forward thrust provided by the airplane. Given that the propellers provide a net forward thrust of 3.6X10^4 N, and the glider's mass is .6X10^4 kg, we can calculate the glider's acceleration using F = ma, where the mass is .6X10^4 kg and the force is 3.6X10^4 N.

To solve these problems, we can apply Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.

1. To find the net force acting on the wagon, we can use the formula: Net force = mass x acceleration.
Given: mass (m) = 20 N, acceleration (a) = 0.15 m/s^2.
Substituting these values into the formula, we get: Net force = 20 N x 0.15 m/s^2 = 3 N.
Therefore, the net force acting on the wagon is 3 N.

2. To find the distance the car travels before stopping, we can use the formula: Net force = mass x acceleration.
Given: mass (m) = 2000 kg, acceleration (a) = -30 m/s^2 (negative as the car is decelerating), and the constant braking force (F) = 10000 N.
Rearranging the formula to solve for distance (d), we get: d = (initial velocity^2) / (2 x acceleration).
Substituting the values into the formula: d = (30 m/s)^2 / (2 x -30 m/s^2) = 900 m^2/s^2 / -60 m/s^2 = -15 m.
The negative sign indicates that the car traveled in the negative direction before stopping, so the car travels 15 meters before stopping.

3. To find the glider's acceleration, we can use the formula: Net force = mass x acceleration.
Given: mass of the airplane (m1) = 1.2 x 10^4 kg, mass of the glider (m2) = 0.6 x 10^4 kg, net forward thrust (F) = 3.6 x 10^4 N.
Since the net force is provided by the airplane's propellers and is equal to the net force acting on the glider, we can equate the two.
Therefore: Net force = 3.6 x 10^4 N = (m1 + m2) x acceleration.
Rearranging the formula to solve for acceleration, we get: acceleration = Net force / (m1 + m2).
Substituting the values into the formula: acceleration = 3.6 x 10^4 N / (1.2 x 10^4 kg + 0.6 x 10^4 kg) = 3.6 x 10^4 N / 1.8 x 10^4 kg = 2 m/s^2.
Therefore, the glider's acceleration is 2 m/s^2.

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