Now we are going to consider how the application of an external force affects an
ID: 1601689 • Letter: N
Question
Now we are going to consider how the application of an external force affects an object's energy and momentum. A 50kg crate is at rest on a frictionless surface. What is the crate's kinetic energy and momentum? (Use K = 1/2mv^2 and p = mv.) Now a 50N force is applied to the right. If this is the only force acting in the x-direction, what is the acceleration of the crate? (Use Newton's 2nd Law.) Calculate the velocity of the crate after this force has been applied for 10s (Use v_f = v_i + at.) Calculate the distance the crate has moved after the force has been applied for 10s. (Use x_f = x_i + v_i t + 1/2 at^2.) Using the final velocity you calculated in step 3, calculate the kinetic energy and momentum of the crate after the force has been applied for 10s. According to the law of conservation of energy and the law of conservation of momentum, a closed system's energy and momentum must be constant. Is this system closed? How do you know? Multiply the applied force by the time the force was applied. Does this value match anything you calculated above? What are the units of this quantity? What other physical quantity has these units? Multiply the force applied by the distance the crate traveled. Does this value match anything you calculated above? What are the units of this quantity? What other physical quantity has these units?Explanation / Answer
1)
crate is at rest
initial velocity vi = 0
kinetic energy Ki = (1/2)*m*0^2 = 0
momentum pi = m*v = 50*0 = 0
(2)
acceleration ax = Fx/m
a = 50/50 = 1 m/s^2
==============
3)
vf = vi + a*t
vf = 0 + (1*10) = 10 m/s
====================
4)
xf = xi + vi*t + (1/2)*a*t^2
xf = 0 + 0 + (1/2)*1*10^2
xf = 50 m
=================
5)
final kinetic energy Kf = (1/2)*m*vf^2 = (1/2)50*10^2 = 2500 J
final momentum Pf =m*vf = 50*10 = 500 kg m/s
=================
NO
there is external force acting
==================
F*t = 50*10 = 500 Ns
It matches with the change in momentum Pf - Pi = 500 - 0 = 500 kg m/s
impulse
===============
F*x = 50*50 = 2500 J
It matches with the change in KE = Kf - Ki = 2500 J
energy
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.