Question

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?

Answer 1

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.

Answer 2

300 m

Explanation:

Displacement in the first 10 seconds:

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

• 24 N = (8 kg) * a

Divide both sides by 8.

• a = 3 m/s²

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

• 3 m/s² = v/10 s

Multiply 10 to both sides.

• v = 30 m/s

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

• v_f = 30 m/s
• a = 3 m/s²
• t = 10 s
• v_i = 0 m/s

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:

• Δx = v_0 t + 1/2at²

Plug in known values into the equation.

• Δx = (0 m/s)(10 s) + 1/2(3 m/s²)(10 s)²

Remove units to make it easier to read.

• Δx = (0)(10) + 1/2(3)(10)²

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

• Δx = 1/2(3)(10)²

Simplify this equation.

• Δx = 1.5 * 100
• Δx = 150 m

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

Displacement in the last 10 seconds:

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:

• v = v_0 + at

Plug in known variables:

• 0 = 30 m/s + a(10 s)

Subtract 30 from both sides of the equation.

• -30 = 10a

Divide both sides of the equation by 10.

• a = -3 m/s²

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

• v_f = 0 m/s
• a = -3 m/s²
• t = 10 s
• v_i = 30 m/s

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

• Δx = v_0 t + 1/2at²

Plug in known values:

• Δx = (30 m/s)(10 s) + 1/2(-3 m/s²)(10 s)²

Remove units to make the equation easier to read.

• Δx = (30)(10) + 1/2(-3)(10)²

Simplify this equation.

• Δx = 300 - 1.5(100)
• Δx = 300 - 150
• Δx = 150 m

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:

• 150 m + 150 m = 300 m

The total distance the object covered is 300 meters.