A 50 kg ice skater is gliding along the ice, heading due north at...

Question

A 50 kg ice skater is gliding along the ice, heading due north at 4.0 m/s. The ice has a small coefficient of static friction, to prevent the skater from slipping sideways, but μₖ = 0. Suddenly, a wind from the northeast exerts a force of 4.0 N on the skater. (a) Use work and energy to find the skater's speed after gliding 100 m in this wind.

Accepted Answer

Everyone in this problem. A skier is skiing in the north direction at 5m/s. The skier experiences a force of three newtons from the northwestern winds. The coefficient of kinetic friction is zero. Although there exists static friction to prevent the skier from skidding sideways. We're asked to determine the skier speed after skiing m along with the wind, We're told to consider the skier has a mass of 45 kg. We're given four answer choices. A is 2.4 m/s. B is 3.5 m/s. C is 4.5 m/s and d is 5.4 m/s. Now, we have our skier, I'm just gonna draw a picture here. So I'm drawing, our skier can give them ski poles, OK? And he is moving directly to the north. They are moving directly to the north at 5m/s. Well, we're gonna take upwards to be our positive direction, our positive y direction and we have the wind and we're told that it's northwestern wind. And when we hear northwestern wind, what that means is the set, the wind is coming from the northwest. So the wind is going to be coming from the northwest, which means that it's actually blowing in the southeast direction. So this is our wind pointing in the southeast direction. Now, in the X direction, we have no initial speed, ok. The skier is skiing directly north. So we have no X direction for the initial speed. And we're told that we have static friction that prevents the skier from skidding sideways. What that means is that we're not gonna have any final speed in the X direction either because the friction is going to prevent that from happening. OK. So we don't need to worry about the X direction. We're gonna have just a Y component of this speed. Now, we have information about this force from the wind. We're trying to find information about speed. So let's recall our work. Kinetic energy theorem tells us that the work W is equal to the change in kinetic energy delta K. What is the work? Well, I recall that we can write the work as the force F multiplied by the displacement D multiplied by cosine of five where Phi is the angle between that fork F and the displacement D and this is gonna equal the change in kinetic energy. Recall that the kinetic energy is given by one half multiplied by M multiplied by V squared. And so the change in kinetic energy is going to be one half multiplied by M multiplied by V F squared is one half multiplied by M multiplied by V not squared. Now, we know the force, we're giving them the problem. We know the distance or the displacement. What about this angle by, let's draw a little diagram to help us with that. But we know that the skier is going straight north. So their displacement vector D is pointing to the north. The force vector is pointing to the southeast and, and this is the force of the wind. We draw the horizontal. We know that the force of the wind makes a 45° angle Because it is a northwestern wind. That means it's rate 45° between north and west. So that's 45 degrees. And so this angle thigh, I draw in red between our displacement vector And the force of the wind is going to be 90° plus that 45°, which is equal to 135°. OK. So we have a 135° angle between those two forces. So that is our value of five. We know the mass and we're told the initial speed as well. So we have everything we need to solve for this final velocity V F or the final speed of the skier. So the four F 3 noons Multiplied by the displacement, 50 m multiplied by cosine of our angle P 5, 135 degrees is equal to one half multiplied by 45 kg multiplied by V F squared. Subtract one half Multiplied by 45 kg, Multiplied by 5m/s squared. Right. If we simplify, on the left hand side, we get negative .06, 6 kilogram meters squared per second squared. And we got Newton which is a kilogram meter per second squared times meter. So we get kilogram meter squared per second squared. On the right hand side, we have 22.5 kg multiplied by V F squared -562. kilogram meter squared per second squared. And we can add that 562.5 kg to both sides. OK? To try to isolate V F. So we get . kilogram meter squared per second squared is equal to 22.5 kg times V F squared, dividing both sides by 22.5 kg. V X squared is equal to 20. m squared per second squared. We can take the square root to get that value of V F. When we take the square root, we have to remember that we get both the positive and the negative root. In this case, we're just asked to find the speed. So we just want the magnitude of that number and so the positive route is fine. And so we have 4. 04 m/s and that is the value we were looking for. So if we go back up to our answer traces, we can see that the answer choices are rounded to two significant digits. We found that the Kear speed is 4.5 m/s, which corresponds with answer choice. C thanks everyone for watching. I hope this video helped see you in the next one.

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