An 8.0 kg object starts from rest and is pushed with a horizontal 24.N force for 10. s on a surface with negligible friction. Then the applied force is removed and the object then encounters a rough surface and slows to a stop in an additional 10. seconds. a) What TOTAL distance did the object cover?

The 24 N force applies an acceleration a₁ such that

24 N = (8.0 kg) a₁

a₁ = (24 N) / (8.0 kg)

a₁ = 3.0 m/s²

Starting from rest, the object accelerates at this rate for 10 s, so that it reaches a velocity v of

v = (3.0 m/s²) (10 s)

v = 30 m/s

and in these 10 s, the object travels a distance d₁ of

d₁ = 1/2 (3.0 m/s²) (10 s)²

d₁ = 150 m

When it hits the rough surface, it slows to a stop in 10 s, so that its accleration a₂ during this time is such that

0 = 30 m/s + a₂ (10 s)

a₂ = - (30 m/s) / (10 s)

a₂ = -3.0 m/s²

so that it covers an additional distance d₂ such that

d₂ = 1/2 (-3.0 m/s²) (10 s)²

d₂ = 150 m

So the object travels a total distance of d₁ + d₂ = 300 m.

Alternatively, once we find the accelerations during both 10 s intervals and the velocity after the first 10 s, we get

(30 m/s)² - 0² = 2 (3.0 m/s²) d₁

d₁ = (30 m/s)² / (6.0 m/s²)

d₁ = 150 m

and

0²- (30 m/s)² = 2 (-3.0 m/s²) d₂

d₂ = (30 m/s)² / (6.0 m/s²)

d₂ = 150 m

so that, again, d₁ + d₂ = 300 m.

300 m

Explanation:  

We can find the acceleration of the object by using Newton's 2nd Law: F = ma.

Divide both sides by 8.

We can find the final velocity (v) using the acceleration since a = v/t.

Multiply 10 to both sides.

List out the variables we have for the object during its first 10 seconds in motion.

We want to solve for the displacement in the x-direction during the first 10 seconds, so we can use this constant acceleration kinematics equation:

Plug in known values into the equation.

Remove units to make it easier to read.

0 * 10 cancels out, and we are left with:

Simplify this equation.

The displacement in the x-direction is 150 m during the first 10 seconds of the object's motion.

Now we want to solve for the acceleration during the last 10 seconds while the object is slowing down and stopping.

Since the object stops, the final velocity is 0 m/s. We know the final velocity of the first 10 seconds is the initial velocity of the last 10 seconds.

Let's use this constant acceleration equation to solve for a:

Plug in known variables:

Subtract 30 from both sides of the equation.

Divide both sides of the equation by 10.

List out known variables for the last 10 seconds of the object in motion:

Let's use the same equation we used for the first 10 seconds since we have the same known variables:

Plug in known values:

Remove units to make the equation easier to read.

Simplify this equation.

The displacement in the x-direction for the last 10 seconds is equal to the displacement for the first 10 seconds.

Let's add these together:

The total distance the object covered is 300 meters.

v=ir = 3x15 =45 v