Energy Conservation and Kinetic, Wind Energies

AI Thread Summary
Jane, weighing 49.0 kg, must swing across a river against a constant wind force of 126 N to rescue Tarzan, who weighs 80.0 kg. To determine the minimum speed for Jane's swing, the potential energy associated with the wind force and the gravitational potential energy must be balanced with the kinetic energy. Calculations reveal varying results for the required speed, indicating confusion over the energy dynamics involved, particularly regarding the height change during the swing. For the return swing with both Jane and Tarzan, the combined mass alters the minimum speed needed to overcome the wind resistance. Accurate calculations are essential to ensure both can safely cross the river.
Yoshistr
Messages
1
Reaction score
0
1. Jane, whose mass is 49.0 kg, needs to swing across a river filled with man-eating crocodiles in order to rescue Tarzan, whose mass is 80.0 kg. However, she must swing into a constant horizontal wind force F on a vine that is initially at an angle of with the vertical. (See Fig. P5.73.) In the figure, D = 50.0 m, F = 126 N, L = 40 m, and theta = 50.0°.

Figure P5.73
http://www.webassign.net/sercp/p5-73.gif

(a) With what minimum speed must Jane begin her swing in order to just make it to the other side? (Hint: First determine the potential energy that can be associated with the wind force. Because the wind force is constant, use an analogy with the constant gravitational force.)

(b) Once the rescue is complete, Tarzan and Jane must swing back across the river. With what minimum speed must they begin their swing?





2. Homework Equations
W = KE + PE
KE = 0.5mv^2
PE = mgh
W = Fcos(theta)*d
Ei = Ef




3.
I first used:
Ei = Ef
PE(gravity) + KE(motion) = W(wind)
mgh + 0.5mv^2 = F(wind)*cos(theta)*d
49*9.81*(40-40*sin40) + 0.5*49*v^2 = 126*cos(50)*50
v = (2*((126*cos(50)*50)-(49*9.81*(40-40*sin40)))/49)^0.5
I got a number of different v:
16.0356
12.856
4.81637
7.7536
5.98
using different formula and substitutions but they are ALL incorrect by at least 10%!

Line of reason:
Jane needs to move against a force field of wind which exertes 126N, when multiplied by 50m the distance the work needed to go against the wind is 126*50 J multiplied by an angle of 50 degrees? Here is where it gets confusing, if she is moving in an arc wouldn't the energy of the wind change also? W = Fcos(theta)*d
So she obviously needs a KE equal to W in order to overcome it, also in the diagram I am unsure if there is a height change, but if there is it is negative energy with respect to the y axis? Should I break up the question into an x and y part then say v = ((vx)^2 + (vy)^2)^0.5?
height difference between starting point and ending point:
h = L - Lsin40
 
Physics news on Phys.org
= 40 - 40sin40 = 25.624(a) With what minimum speed must Jane begin her swing in order to just make it to the other side? Ei = EfPE(gravity) + KE(motion) = W(wind)mgh + 0.5mv^2 = F(wind)*cos(theta)*d49*9.81*25.624 + 0.5*49*v^2 = 126*cos(50)*50v = (2*((126*cos(50)*50)-(49*9.81*25.624))/49)^0.5v = 16.0356 m/s(b) Once the rescue is complete, Tarzan and Jane must swing back across the river. With what minimum speed must they begin their swing?Ei = EfPE(gravity) + KE(motion) = W(wind)mgh + 0.5mv^2 = F(wind)*cos(theta)*d129*9.81*25.624 + 0.5*129*v^2 = 126*cos(50)*50v = (2*((126*cos(50)*50)-(129*9.81*25.624))/129)^0.5v = 12.856 m/s
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top