Jason takes off from rest across level water on his jet-powered skis. The combined mass of Jason and his skis is 75 kg (the mass of the fuel is negligible). The skis have a thrust of 200 N and a coefficient of kinetic friction on water of 0.10. Unfortunately, the skis run out of fuel after only 75 s. What is Jason's top speed?

Answers

Answer 1

Answer:

 v = 126 m / s

Explanation:

Let's analyze this exercise a little, they give us the thrust that is the applied force and the time that it lasts, and they ask us for the final speed, so we can use the Impulse ratio and the variation of the amount of movement

       I = F t = Dp

       F t = pf -p₀

      Now let's use Newton's second law to find the net thrust

        F = E - fr

the friction force has the formula

       fr = μ N

let's write Newton's second law on the y-axis

       N-W = 0

       N = W

we substitute

      fr = μ mg

we look for the net out

       F = 200 - μ mg

With the skater starting from rest, the initial speed is zero (vo = 0)

we substitute

      (200 - very m g) t = m v

       v = (200 µm - very g) t

let's calculate

     v = (200/75 - 0.10 9.8) 75

      v = 126 m / s


Related Questions

Two workers are sliding 330 kg crate across the floor. One worker pushes forward on the crate with a force of 430 N while the other pulls in the same direction with a force of 330 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor?

Answers

Answer:

Coefficient of kinetic friction = 0.235

Explanation:

Given:

Mass of crate = 330 kg

1st force = 430 N

2nd force = 330 N

Find:

Coefficient of kinetic friction.

Computation:

We know that, velocity is constant.

So, acceleration (a) = 0

So, net force (f) = 430 N + 330 N

Net force (f) = 760 N

F = μmg

μ = f / mg                                   [∵ g = 9.8]

μ = 760 / [330 × 9.8]

μ = 760 / [3,234]

μ = 0.235

Coefficient of kinetic friction = 0.235

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