What Force Maintains a Constant Velocity for a Water Skier?

AI Thread Summary
To maintain a constant velocity for a water skier, the pulling force must equal the total resistive force. The skier experiences a horizontal force of 495 N and an acceleration of 2.0 m/s², leading to a net force calculation of 150 N. However, the correct approach involves recognizing that the net force is the difference between the pulling force and the resistive forces. Therefore, to find the force needed for constant velocity, one must account for both the pulling force and the resistive forces acting against it. Understanding these dynamics is crucial for solving the physics problem accurately.
shawonna23
Messages
146
Reaction score
0
Physics Homework Problem--Stuck?

A 75 kg water skier is being pulled by a horizontal force of 495 N and has an acceleration of 2.0 m/s2. Assuming that the total resistive force exerted on the skier by the water and the wind is constant, what force is needed to pull the skier at a constant velocity?


I tried doing this to solve the problem:

F=ma
F=75kg x 2.0= 150N

Then I added 150 and 495 to get 645 N but this is not the right answer.

Can someone please tell me what I did wrong?
 
Physics news on Phys.org
What is the total resistive force...?

Daniel.
 
F_{net} = ma

This stands for the net force.. not only one. Draw a force diagram and see where each one is going.
 

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
View full post »
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

Total resistive force by water and air on water skier
Replies
1
Views
3K
What Are the Gravitational and Normal Forces on a Skier on a Slope?
Replies
4
Views
2K
Terminal velocity of a skier using a Momentum Balance
Replies
1
Views
2K
Why must a water skier at constant velocity lean back?
Replies
13
Views
5K
Newton's Second Law - Net Force and Constant Acceleration
Replies
6
Views
3K
How to interpret the given informations in physics problems
Replies
16
Views
2K
Two carts connected with a spring -- working with forces and acceleration
Replies
14
Views
3K
Solving Fx for 50kg Skier Down Hill
Replies
2
Views
3K
How to calculate the wind force acting on a water drop?
Replies
39
Views
3K
Physics Question involving a skier and a snowdrift....
Replies
5
Views
3K

Hot Threads

Recent Insights

Back
Top