- #1

imdumb

- 3

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

In summary: So the spring constant k = 30 [N/m].In summary, a slingshot is made up of a leather cup with a stone and two parallel rubber bands. It takes a force of 15 N to stretch one of the bands 1.0 cm. With a 47g stone pulled back.12 m from equilibrium, the potential energy in the two bands combined is 41.58 J. The speed of the stone leaving the slingshot can be calculated using the equation PE = .5kx^2, with k = 30 [N/m] and x = .12 m, resulting in a speed of 4.08 m/s.

- #1

imdumb

- 3

- 0

Physics news on Phys.org

- #2

quasar987

Science Advisor

Homework Helper

Gold Member

- 4,807

- 32

you have to consider the rubber band has happy Hookian springs. I.e. such that F = kx.

By "It takes a force of 15 N to stretch either one of the bands 1.0 cm", what they mean is that if you exert a force of 15N on one of the rubber band, you're not going to be able to stretch it farther than 1cm. At this point, equilibrium will have been attained. this means that at x = 1cm, the force of the rubber band is of 15N too. From this, you get k = 15 [N/cm].

You go from there.

By "It takes a force of 15 N to stretch either one of the bands 1.0 cm", what they mean is that if you exert a force of 15N on one of the rubber band, you're not going to be able to stretch it farther than 1cm. At this point, equilibrium will have been attained. this means that at x = 1cm, the force of the rubber band is of 15N too. From this, you get k = 15 [N/cm].

You go from there.

Last edited:

- #3

imdumb

- 3

- 0

so... then would 7.5 be the correctanswer to part A?

- #4

quasar987

Science Advisor

Homework Helper

Gold Member

- 4,807

- 32

Nope. How did you get that answer?

(Btw - I fixed some dimension errors in my first post. I replaced all meters with centimeters)

(Btw - I fixed some dimension errors in my first post. I replaced all meters with centimeters)

Last edited:

- #5

imdumb

- 3

- 0

I used the equation PE=.5kx^2and put 15 infor k and 1 in for x.

- #6

quasar987

Science Advisor

Homework Helper

Gold Member

- 4,807

- 32

The actual problem talks about a rock that is pulled back.12 m (12 cm, not 1cm). Moreover, 15 [N/m] is the spring constant of ONE rubber band. You got two.

Speed is a measure of how fast an object is moving, typically measured in meters per second (m/s) or kilometers per hour (km/h).

The speed of the stone leaving the slingshot is affected by the tension of the slingshot, the angle at which it is released, the weight and size of the stone, and any external forces such as wind resistance.

The speed of the stone can be calculated using the equation v = √(2gh), where v is the speed, g is the acceleration due to gravity, and h is the height from which the stone is released. This assumes no air resistance and a horizontal release.

Yes, the speed of the stone will change as it travels through the air due to the effects of air resistance. The stone will slow down as it moves through the air, and its speed will depend on its weight, shape, and the surrounding air conditions.

Yes, the speed of the stone can be increased by increasing the tension of the slingshot, using a heavier or smaller stone, or releasing the stone at a higher angle. However, there will be a limit to how fast the stone can travel due to external factors like air resistance and the strength of the slingshot.

- Replies
- 3

- Views
- 16K

- Replies
- 2

- Views
- 4K

- Replies
- 4

- Views
- 13K

- Replies
- 5

- Views
- 2K

- Replies
- 7

- Views
- 9K

- Replies
- 2

- Views
- 21K

- Replies
- 4

- Views
- 2K

- Replies
- 14

- Views
- 3K

- Replies
- 15

- Views
- 6K

- Replies
- 16

- Views
- 4K

Share: