Dynamics and Forces: Solving for Safety Line Tightness and Friction Force

  • Thread starter Thread starter Omelette
  • Start date Start date
  • Tags Tags
    Dynamics Forces
AI Thread Summary
The discussion focuses on solving two physics problems involving dynamics and forces. For the first question, the astronaut pushes off the spacecraft, and the time for the safety line to become tight is calculated using conservation of momentum, leading to a corrected answer of 2.34 seconds after addressing the relative velocities. The second question involves determining the friction force acting on a 3.0 kg object being pulled by a 2.0 N force while accelerating at 1.2 m/s². The user initially struggled with the calculations but received guidance on correctly applying the equations of motion and understanding net forces. Overall, the thread emphasizes the importance of correctly applying physics principles to solve dynamics problems.
Omelette
Messages
3
Reaction score
0

Homework Statement



Q1) After a 1.30 x 10^2kg astronaut (incl. equipment) connects a safety line (length = 22.0m) to herself and to the spacecraft (mass = 2.80 x 10^3kg), she pushes against the craft and moves away at a constant velocity of 9.0m/s. How long does it take the safety line to become tight?

Q2) A 3.0kg object is pulled along a horizontal surface as shown in the diagram below by a horizontal force of 2.0N. If the object is accelerating at a rate of 1.2m/s^2, what is the force of friction acting on the object? (Diagram below).

Homework Equations



Fnet = ma
d = Vot + .5at^2
a = (Vf - Vo)/2
d = (Vf + Vo)/2 * t
Vf^2 = Vo^2 + 2ad
Ff = μFn

The Attempt at a Solution



Q1) I thought to set d = 22, Vo = 0, Vf = 9, and solve for t using the 4th equation, but that answer isn't one of the choices. I think maybe the masses of the objects need to be used, but I don't know how they may help.

Q2) I used F = ma, and so Fnet = (3)(1.2) = 5.2 N (which is also on the diagram). I thought about Fnet = Fapp - Ff, and tried to solve for Ff by subtracting Fnet from Fapp, but that isn't one of the choices as well.

I don't really know where to start with these questions.
Help would be much appreciated! Thanks!
 

Attachments

  • phys.png
    phys.png
    633 bytes · Views: 412
Physics news on Phys.org
Initially astronaut are at rest with respect to each other. After the push astronaut moves with the velocity va. Using the conservation of momentum, find the velocity of the spacecraft vs. Both are moving in the opposite. The relative velocity between them is (va+vs). So time to move a distance d is t = d/(va+vs)
 
Using the conservation of momentum, I found the velocity of the spaceship to be v = p/m --> v = -1170/2800 = -0.41786m/s. Then I solved for t by doing t = d/(va + vs) = 22/(9 - 0.41786) = 2.56s, but the multiple choice answers are either 0.418s, 2.34s, 0.900s, or 2.44s.

Does anyone know how I should go about solving Q2?

Thanks again for your help!
 
22/(9 - 0.41786)
It should be 22/(9 + 0.41786) because when the two particles are moving in the opposite direction, their relative velocity is the sum of their velocities.
 
Ohh, I see that now. Thank you! I found the answer to be 2.34s.
 
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 »

Thread 'Struggle to understand the concept of the 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 »

Similar threads

Dynamics ramp and friction. Finding angle?
Replies
4
Views
2K
How Do I Determine Force of Static Friction on a Rolling Car
Replies
15
Views
2K
Finding Force Applied on a ball from another object
Replies
1
Views
1K
Coefficient of kinetic friction forces
Replies
3
Views
2K
Forces and Free-Body Diagrams in Real-World Scenarios
Replies
5
Views
4K
Determining Coefficient of Friction for Two Boxes with Applied Force
Replies
14
Views
2K
Action-Reaction Forces on an Elevator
Replies
5
Views
3K
Basic Physics Ramp & Friction Question
Replies
1
Views
2K
Finding an unknown applied force with friction, Fnet and more.
Replies
2
Views
3K
Dynamics Question: Finding mass using Vi, Vf, D and acceleration
Replies
6
Views
4K

Hot Threads

Recent Insights

Back
Top