Question

A slingshot is a child's toy that converts the
strain energy ofthe stretched material into
the moving energy of the propelled object. A
boy pulls on his slingshot with a force of 20N
and it extends by 7 cm. At what velocity
does the 20 g stone fly off?
(A) 8.4 m S-1
(B) 70ms-1
(e) 6.7 m s-1
(D) 2.6 m s-1

Answer 1

Do you know the work energy theorem? It says that the amount of work you put into a system (minus losses) is contained in that system, available as work. Work is defined as Force times distance (Fd). That tells you how much potential energy the boy put in his slingshot. That will be converted to kinetic energy\[K=(1/2)mv ^{2}\]
Set K equal to Fd and solve for v. Dont forget to convert cm to m!

Answer 2

Hint: Work transferred to some object is given by the following line integral.\[W=\int\vec F \cdot d\vec x\]When the boy stretches the slingshot, he does work on the bands of the slingshot, which is equivalent to saying that he transfers energy to the slingshot. When it is released, this energy that he transfers is released into the kinetic energy. Equate the value for work with the expression for kinetic energy \(K\) given by egenriether above and you will find the final velocity after you solve for \(v\) and substitute in the given values.

Answer 3

could you just say which answer is correct

Answer 4

Unless I've made a mistake, I'm getting 11.83 m/s...which isn't one of the listed answers. Anyone else care to check my math?

Answer 5

that's what i got too -_- which is why im puzzled :/

Answer 6

Let's take a different approach than the work-energy theorem. Let's use conservation of energy.

The potential energy stored in a spring, is a function of the spring constant and the square of the distance displaced, \[E_s = {1 \over 2} kx^2\]. We can find the spring constant from Hooks Law\[F_s = kx \rightarrow 20 = k \cdot 0.07\]

Now, from conservation of energy, \[{1 \over 2} m v^2 = {1 \over 2} kx^2\]

This technique yields an answer that is listed.

Answer 7

i got 8.36 m/s which is one of the listed answers!

Answer 8

Indeed.

The discrepancy in the use of the Work-Energy theorem arises from the fact that a spring force is not a constant force. This means that\[W_s \ne F_s \cdot x\]Instead, observe that\[W = \int\limits F dx\]\[W_s = \int\limits kx dx = {1 \over 2} k x^2\]

Answer 9

The use of the Work-Energy Theorem is valid here. Just make sure you are using the correct expression for work.

Answer 10

thanks a lot!

Answer 11

You are welcome!

Answer 12

Good catch @eashmore....<smacks forehead>...I should have realized we were using the wrong work formula :)