Velocity from Kinetic energy and work energy theorem

AI Thread Summary
To determine the speed of a 0.066 kg arrow fired horizontally with a bowstring force of 50 N over a distance of 0.95 m, the work-energy theorem is applied. The work done on the arrow is calculated as the force multiplied by the distance, resulting in 47.5 J. This work is equal to the change in kinetic energy of the arrow, which is expressed as (1/2 * mass * Final Velocity^2). Since the initial velocity is zero, the equation simplifies to Final Velocity = Square root (2 * Work / mass). The correct calculation reveals the arrow's exit speed from the bow.
IAmSparticus
Messages
36
Reaction score
0
1. A 0.066 kg arrow is fired horizontally. The bowstring exerts an average force of 50 N on the arrow over a distance of 0.95 m. With what speed does the arrow leave the bow?


2. Work Energy Theorem = change in kinetic energy = (1/2*mass*Final Velocity^2)-(1/2*mass*Initial Velocity^2)



3. Since the initial speed is zero and the mass is given, I get a solution of 0, but that is most likely because I did the algebra wrong. I got an equation of Final Velocity = Square root (2*.066kg*0m/s)/(2*.066)

Which is clearly wrong. Where did I go wrong?
 
Physics news on Phys.org
What is the work done on the arrow?
Equate it to the kinetic energy of the arrow and find the velocity.
 
Thread 'Collision of a bullet on a rod-string system: query'

Thread 'Collision of a bullet on a rod-string system: query'

In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
View full post »
Thread 'A cylinder connected to a hanged mass'

Thread 'A cylinder connected to a hanged mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Conservation of energy of a bow and arrow
Replies
3
Views
761
A block dropping onto a spring system
2
Replies
58
Views
2K
Resistance force changing the velocity of an object
Replies
8
Views
1K
Work energy KE theorem for a book being lifted up in a gravitional field
Replies
18
Views
2K
Questions Regarding a Cat Going Up a Ramp
2
Replies
56
Views
3K
Understanding the Work-Energy Theorem: Solving the Roller Coaster Problem
Replies
8
Views
3K
Applying the Work-Energy theorem to a system
Replies
9
Views
1K
Difficulty in deciding when to apply work energy theorem
Replies
2
Views
1K
Internal energy of a gas and kinetic energy, "typical velocity"
Replies
5
Views
754
Understanding Work-Energy Theorem: Solving an Exercise with Different Solutions
Replies
11
Views
2K

Hot Threads

Recent Insights

Back
Top