The most powerful tugboats in the world are built in finland. theseboats exert a force with a magnitude of 2.85× 10^6n. suppose one ofthese tugboats is trying to slow a huge barge that has a mass of 2.0× 10^7kg and is moving with a speed of 3.0 m/s. if the tugboat exerts its maximum force for 21 s in the direction opposite to that in which thebarge is moving, what will be the change in the barge’s momentum? howfar will the barge travel before it is brought to a stop?

-5.985*10^7Ns=momentum

ds=31.578m

Explanation:

The most powerful tugboats in the world are built in Finland. Theseboats exert a force with a magnitude of 2.85× 10^6N. Suppose one ofthese tugboats is trying to slow a huge barge that has a mass of 2.0× 10^7kg and is moving with a speed of 3.0 m/s. If the tugboat exerts its maximum force for 21 s in the direction opposite to that in which thebarge is moving, what will be the change in the barge’s momentum? Howfar will the barge travel before it is brought to a stop?

from newton's second law of motion , which states that the rate of change in momentum is directly proportional to the force applied

f=kdv/dt

k=1

ft=dv/dt

dp=dv/dt

f=negative because its moving in the opposite direction

2.85× 10^6*21=

-5.985*10^7Ns=momentum

-dp=mv2-mv1

-dp+mv1=mv2

-5.985*10^7+2.0× 10^7 kg*3.0 m/s=2.0× 10^7kgV2

v2=150000/(2.0× 10^7)

v2=0.0075m/s to the right

ds=1/2(v2+v1)dt

ds=0.5*(3.0075)*21

ds=31.578m

the barge will travel 31.578m before it is brought to rest.

∆p = –5.985×10⁷ kg×(m/s)

∆x = 31.57875 m

The boat's change in momentum is –5.985×10⁷ kg×(m/s), and it will travel 31.57875 meters before it stops.

Explanation:

First off, let's take a quick look at the formula for momentum: momentum equals the force times the time the force is applied for. This means:

p = FΔt

Why is momentum represented by the letter p? That's actually a good question, to be honest—I guess they were running out of letters because the letter m is already used for "mass". But I digress.

Let's go back to the equation. We have both those values—the force applied and the time it is applied for—and so we can substitute those into the equation. But be careful: here, the force is applied in the opposite direction as the actual motion. This means the force must be negative. We get:

p = (–2.85×10⁶ N)(21 s)

p = –5.985×10⁷ kg·(m/s)

That's our value for initial momentum, not the final momentum. Why? Momentum and impulse are equivalent and their units are the same. If we calculated this as the impulse (J = mΔv), we'd be using the initial velocity, not the final velocity (the velocity after the force is applied).

But this problem asked us to solve for two things: the change in momentum (Δp) and the stopping distance (Δx). We can't find either value if we don't know the velocity of the block after the force is applied. This means we need to solve for it.

This is where knowing impulse and momentum are equal comes in handy: if the two values are equal, their formulas should be equal, too. It's like solving a system of equations with the substitution method. With the impulse and momentum formulas, we make them equivalent and get:

mΔv = FΔt

The change in something is like the difference in subtraction: it's the final value minus the initial value. We can then rewrite the equation as:

m(v₂ – v₁) = FΔt

Here, I used (v₂) for the final (second) velocity and (v₁) for the initial (first) velocity. Now, since we need to find the final velocity, (v₂), we need to isolate it by solving the equation for this value.

m(v₂ – v₁) = FΔt

[m(v₂ – v₁)] ÷ m = (FΔt) ÷ m

v₂ – v₁ = (FΔt) ÷ m

v₂ – v₁ + v₁ = [(FΔt) ÷ m] + v₁

v₂ = [(FΔt) ÷ m] + v₁

That's our formula to find the final velocity. The problem already gave us all the values we need to solve this equation: the force applied, the time the force is applied for, the mass of the object being stopped, and the object's initial velocity. When we substitute those into the equation, we get:

v₂ = [(FΔt) ÷ m] + v₁

v₂ = [(–2.85×10⁶ N)(21 s) ÷ 2.0×10⁷ kg] + 3.0 (m/s)

v₂ = [(–5.985×10⁷ kg·(m/s)) ÷ 2.0×10⁷ kg] + 3.0 (m/s)

v₂ = 3.0 (m/s) – 2.9925 (m/s)

v₂ = 0.0075 (m/s)

That gives us the final velocity that we need to find (Δp) and (Δx). Now, using the equations for each value, we can substitute in and finish this up!

Δp = m(Δv) = m(v₂ – v₁)

Δp = (2.0×10⁷ kg)[0.0075 (m/s) – 3.0 (m/s)]

Δp = (2.0×10⁷ kg)[–2.9925 (m/s)]

Δp = –5.985×10⁷ kg·(m/s)

Δx = ¹/₂(v₂ + v₁)(Δt)

Δx = ¹/₂[0.0075 (m/s) + 3.0 (m/s)](21 s)

Δx = ¹/₂[3.0075 (m/s)](21 s)

Δx = [3.0075 (m/s)](10.5 s)

Δx = 3.157875 m

And there we go! Problem solved! I hope this helps you! Have a great day!

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.