How Fast Does a Stone Shoot from a Rubber-Band Slingshot?

In summary, the problem involves finding the initial speed of a 25-g stone shot from a rubber-band slingshot, with a draw back distance of 0.15m and a force of 27N. Using the equations for potential and kinetic energy, the initial speed can be found by setting the force equal to the kinetic energy and solving for v. However, in the given attempt at a solution, the mass was incorrectly calculated as 0.25kg instead of 0.025kg, leading to an incorrect answer of 5.69m/s. The correct initial speed should be 11.38m/s.
  • #1
XxphysicsxX
22
0
the problem is:
A rubber-band slingshot shoots a 25-gstone. What is the initial speed of the stone if the rubberband is drawnback 0.15m with aforce of 27 N?
m= 25g
d=0.15 m
f= 27 N
v(initial) = ?


Homework Equations


Ep = (1/2)Fx
Ek=(1/2) mv^2


The Attempt at a Solution


so far, I turned the forceto energy: (27N)(0.15m)= 4.05J
and then solved for v, using Ek= (1/2)mv^2
and I get v=5.69
Im so confused, this answer deffinetly does not seem right..
Im just not sure what to do with the given force here!, do I turn it into work ? (w=fd)
I've been stuck on this question for two days now,i just don't understand how any of the givens can tie in with the energy formulas..?
 
Last edited:
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  • #2
Your method was correct, however your calculation for the value of v is incorrect, as it is too low. You should get a larger answer.
 
  • #3
Well, I am using the exact numbers given ,just converting grams to kg, ...
Ek= (1/2)mv^2
4.05J= (1/2)(0.25)V^2
4.05J/0.125 =V^2
V=5.69

Where am I making a mistake?? :s
 
  • #4
XxphysicsxX said:
Well, I am using the exact numbers given ,just converting grams to kg, ...
Ek= (1/2)mv^2
4.05J= (1/2)(0.25)V^2
4.05J/0.125 =V^2
V=5.69

Where am I making a mistake?? :s

Double check the mass.
 
  • #5
Woww, I feel stupid now, haha,

Thankyou!:biggrin:!
 

1. What is work and energy in physics?

Work and energy are fundamental concepts in physics that describe the ability of a system to do work and the measure of the total amount of energy within a system. Work is defined as the product of the force applied to an object and the displacement of the object in the direction of the force. Energy is the ability of a system to do work, and it exists in different forms such as kinetic, potential, and thermal energy.

2. What is the formula for calculating work?

The formula for calculating work is W = F * d * cosθ, where W is the work done, F is the applied force, d is the displacement of the object, and θ is the angle between the force and the direction of motion. This formula is known as the work-energy theorem and is used to calculate the amount of work done on an object.

3. How is energy conserved in a system?

The law of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. This means that the total amount of energy in a closed system remains constant. In other words, the energy at the beginning of a process will be equal to the energy at the end of the process.

4. What is the difference between kinetic and potential energy?

Kinetic energy is the energy an object possesses due to its motion, while potential energy is the energy an object possesses due to its position or state. Kinetic energy is directly proportional to the mass and velocity of an object, while potential energy is directly proportional to the mass and height of an object.

5. How is power related to work and energy?

Power is the rate at which work is done or energy is transferred. It is calculated by dividing the work done by the time taken to do the work. The unit of power is watt (W), which is equal to one joule (J) per second (s). In other words, power is a measure of how quickly energy is transformed or transferred.

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