Kinetic Energy and Time Problem

AI Thread Summary
A ball projected horizontally from a table has an initial kinetic energy K and gains kinetic energy to 3K after time t, neglecting air resistance. The problem requires relating time to kinetic energy through projectile motion equations. The conservation of energy principle indicates that the ball gains an additional 2K by losing potential energy as it falls. To find time, one must calculate how long it takes to fall a distance h with zero initial vertical velocity. The final answer for time t is (2/g)(√K/m).
jtim36
Messages
2
Reaction score
0

Homework Statement


A ball with mass m projected horizontally o the end of a table with an initial kinetic energy K. At a time t after it leaves the end of the table it has kinetic energy 3K. What is t? Neglect air resistance.
(answer: (2/g)(√K/m)

Homework Equations


E = 1/2 mv^2


The Attempt at a Solution


I understand what the problem is asking, but I don't get how to relate time and kinetic energy
 
Physics news on Phys.org
What equations do you know for the motion of a projectile under gravity?
 
x = vo(t) + 1/2at^2
v^2 = v0^2 + 2ah

Okay, so does that mean the final velocity was tripled then? and I apply the above equations?
 
jtim36 said:
x = vo(t) + 1/2at^2
v^2 = v0^2 + 2ah

Okay, so does that mean the final velocity was tripled then? and I apply the above equations?

No. Not tripled. Think conservation of energy. You have to gain energy 2K by losing potential energy mgh. Put time into it by figuring how long it takes to fall a distance h with zero initial vertical velocity.
 
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