# Dynamics & Forces Questions

1. Jan 30, 2010

### Omelette

1. The problem statement, all variables and given/known data

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).

2. Relevant equations

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

3. 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.

Help would be much appreciated! Thanks!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

• ###### phys.png
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2. Jan 30, 2010

### rl.bhat

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)

3. Jan 31, 2010

### Omelette

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?

4. Jan 31, 2010

### rl.bhat

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.

5. Jan 31, 2010

### Omelette

Ohh, I see that now. Thank you! I found the answer to be 2.34s.