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A rope swing is hung from a tree right at the edge of a small creek. The rope is 5.0 m long; the creek is 3.0 m wide.

1. You sit on the swing, and your friend gives you a gentle push so that you swing out over the creek. How long will it be until you swing back to where you started?
A. 4.5 s
B. 3.4 s
C. 2.2 s
D. 1.1 s

2. Now you switch places with your friend, who has twice your mass. You give your friend a gentle push so that he swings out over the creek. How long will it be until he swings back to where he started?
A. 4.5 s
B. 3.4 s
C. 2.2 s
D. 1.1 s

3. Your friend now pushes you over and over, so that you swing higher and higher. At some point you are swinging all the way across the creek—at the top point of your arc you are right above the opposite side. How fast are you moving when you get back to the lowest point of your arc?
A. 6.3 m/s
B. 5.4 m/s
C. 4.4 m/s
D. 3.1 m/s

Respuesta :

jolis1796

Answer:

1) T = 4.5 s

2) T = 4.5 s

3) v = 9.9 m/s

Explanation:

We can use the equation

T = 2π√(L/g)

1) T = 2π√(5m/9.81 m/s²) = 4.5 s

2) T = 2π√(L/g)

T = 2π√(5m/9.81 m/s²) = 4.5 s

3) v = √(2gR)

v = √(2(9.81 m/s²)(5m))

v = 9.9 m/s

Answer:

2.24

2.24

6.3 m/s

Explanation:

- The given problem can be modeled like a swinging pendulum.

- We will use already derived expressions for SHM of a simple pendulum, so for the time period we have the expression:

                                       T = 2*pi * sqrt ( L / g )

                                       T = 2*pi*sqrt (5.0/9.81)

                                       T = 4.4857 s

- The entire cycle takes 4.4857 s, to complete. However, we are asked to find the half cycle or that is the first time he reaches his starting point.

- Hence, 1/2 T                 t= 2.24 s

- We can see that the time period of the SHM is independent of the mass of the object, hence the answer to second question is also t = 2.24

c)

- In this case we will have to determine the angle that the rope makes with the vertical position:

- We are told its right across the creek, so the angle can be computed:

                                        cos(Q) = 3 / 5

- Now the maximum velocity of a pendulum is given by:

                                        v_max = sqrt(2*g*L*(1-cosQ))

- plug in the values:        v_max = sqrt(2*9.81*5*(1-3/5))

                                        v_max = 6.26 m/s

Hence option A is 6.3 m/s is correct.

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