Help Solving Physics Problems: KE, Velocity & % Increase

  • Thread starter Thread starter thschica
  • Start date Start date
AI Thread Summary
To solve the first problem regarding the arrow's speed, the average force exerted by the bowstring can be used to calculate the work done on the arrow, which equals the change in kinetic energy. The formula KE = 0.5m(v^2) can be applied after determining the work done using the product of force and distance. For the second problem, the kinetic energy at both speeds (94 km/h and 103 km/h) should be calculated, and the percentage increase in kinetic energy can be found by taking the difference in kinetic energies, dividing by the initial kinetic energy, and multiplying by 100. Understanding the relationship between work and energy, as well as Newton's laws, is crucial for solving these types of physics problems. Mastery of these concepts will help in accurately solving similar physics questions.
thschica
Messages
47
Reaction score
0
I am not sure how I am supposed to do these problems and I am going to be tested on them soon can someone please help me.

1) An 83 g arrow is fired from a bow whose string exerts an average force of 95 N on the arrow over a distance of 85 cm. What is the speed of the arrow as it leaves the bow? Do I use KE=.5m(v^2)?. If so how do I use that since I don't know the velocity?

2)An automobile is traveling along a highway at 94 km/h. If it travels instead at 103 km/h, what is the percent increase in the automobile's kinetic energy? I have used KE=.5m(v^2) but I don't understand how I get the percentage!

Please I need some major help!:cry:
 
Physics news on Phys.org
Well, assuming that the arrow starts at rest and then is accelerated by the force of the bowstring, the change in energy is just the integral of force over the distance over which the force is applied. However, since this an average force, one can simply use the product of force * distance, then use the relationship of kinetic energy to velocity and solve for the velocity.


#2. Calculate the KE for each velocity.

Then calculate the difference in the two KE's, i.e. \DeltaKE.

The % difference is just \DeltaKE/KE1*100.

Example, something is traveling at 100 mph when it accelerates to 110 mph.
110 mph represents a 10% increase of 100 mph from (110 - 100)/100 * 100.

100% = 1.
 
I understand #2 but #1 is still very confusing to me and I don't know how to do it.
 
Are you comfortable about the equivalence of Work and Energy.

Remember Newton's Second Law - F = ma.

And remember the kinematic equation which relates acceleration, a, to some displacement, \Deltax,

2a \Deltax = vf2 - vf2.

Multiplying this equation by m and dividing by 2 gives

ma \Deltax = 1/2 mvf2 - 1/2 mvf2, and applying Newton's law,

F\Deltax = KEf - KEi = \DeltaKE.

This of course assumes that F is constant or an average value applied over some distance or displacement, \Deltax.
 
Last edited:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging 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...
Back
Top