Calculating Duck's Displacement in 2.52s with F=ma and Kinematics Equations

  • Thread starter Thread starter scorks
  • Start date Start date
  • Tags Tags
    Force
AI Thread Summary
A duck with a mass of 2.35 kg experiences a force of 0.144 N due east and a current force of 0.202 N at 54.5° south of east, resulting in a net force of 0.261 N. The acceleration calculated using F=ma is approximately 0.11 m/s², leading to a displacement of 0.677 m in 2.52 seconds. Discrepancies in the angle of displacement were noted, with one user calculating an angle of 328.5° relative to due east, while another found a southward distance of about 0.2 m and an angle of less than 20° south of east. The discussion highlights the importance of accurately determining both magnitude and direction in kinematic calculations. Overall, the calculations suggest that the angle may be the source of error in the displacement results.
scorks
Messages
5
Reaction score
0
1.) A duck has a mass of 2.35 kg. As the duck paddles, a force of 0.144 N acts on it in a direction due east. In addition, the current of the water exerts a force of 0.202 N in a direction of 54.5° south of east. When these forces begin to act, the velocity of the duck is 0.130 m/s in a direction due east. Find the magnitude and direction (relative to due east) of the displacement that the duck undergoes in 2.52 s while the forces are acting.

2.) F=ma
Kinematics equations

3.) So, I worked out that the x-component of F2 is 0.117N and added that to 0.144N. I then plugged the answer (0.261N) into the F=ma formula and got a=0.11m/s^2. I then used the equation d=(Vi)t+1/2(a)(t^2) and got d=0.677m.
Then, I used the x-component of 0.261N and the Y-component of F2(the only y-component) and solved for the angle getting 31.5deg. I subtracted this from 360 to get the answer of 328.5deg relative to due east. My answers came out to be incorrect.
Please help!
 
Physics news on Phys.org
For the east direction, I agree with your force of .261 N and a = 0.111.
I used V = Vi + at to get the final velocity of .4099 and then used
d = Vavg *t = (.13+.4099)/2*2.52 = .6803 m East.
Very close to your .677 but could that be what is getting you the wrong answer?
 
Delphi51 said:
For the east direction, I agree with your force of .261 N and a = 0.111.
I used V = Vi + at to get the final velocity of .4099 and then used
d = Vavg *t = (.13+.4099)/2*2.52 = .6803 m East.
Very close to your .677 but could that be what is getting you the wrong answer?

Um, I think that it may be the direction that's getting me the wrong answer. What did you get for that?
 
Finished it now. We are way different on the angle. For the south distance I got about 0.2 m and an angle of less than 20 degrees south of east, which would be more than 340 degrees counterclockwise from east.
 
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}...
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...
Back
Top