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A water-skier with a mass of 68 kg is pulled with a constant force of 870 N by

a speedboat. A wave launches him in such a way that he is temporarily
airborne while still being pulled by the boat, as shown in the image below.
Assuming that air resistance can be ignored, what is the vertical acceleration
that the water-skier experiences on his return to the water surface? (Recall
that g = 9.8 m/s²)
Rope Force
870 N
Weight

User RobbieE
by
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1 Answer

4 votes

To find the vertical acceleration of the water-skier, you can use the formula:

F = ma

where F is the net force acting on the water-skier, m is the mass of the water-skier, and a is the acceleration.

In this case, you know the mass of the water-skier and the force being applied by the rope, so you can solve for the acceleration.

First, you need to find the net force acting on the water-skier. The net force is equal to the sum of all forces acting on the water-skier. In this case, there are two forces acting on the water-skier: the force of gravity and the force being applied by the rope.

The force of gravity is equal to the weight of the water-skier, which is equal to the mass of the water-skier multiplied by the acceleration due to gravity (g):

F_gravity = m * g = 68 kg * 9.8 m/s^2 = 666.4 N

The net force acting on the water-skier is equal to the sum of the forces of gravity and the rope force:

F_net = F_gravity + F_rope = 666.4 N + 870 N = 1536.4 N

Then, you can use the formula for acceleration to solve for the acceleration of the water-skier:

a = F_net / m = 1536.4 N / 68 kg = 22.6 m/s^2

So the vertical acceleration of the water-skier is approximately 22.6 m/s^2.

It's important to note that this calculation assumes that there is no air resistance acting on the water-skier. If there is air resistance present, the actual acceleration of the water-skier may be different.

User Klor
by
7.8k points