An Asteroid is moving along a straight line

AI Thread Summary
An asteroid with a mass of 4.5*10^4 kg is slowing down from 7100 m/s to 5500 m/s due to a force acting along its displacement. To find the work done by the force, the kinetic energy equations can be used, leading to the calculation of initial and final kinetic energies. The change in kinetic energy will provide the work done, which can then be related to the force using the displacement of 1.8*10^6 m. The discussion emphasizes the importance of correctly applying formulas and maintaining unit consistency throughout calculations. Understanding these principles is crucial for accurately solving the problem.
MajesticPenguin
Messages
1
Reaction score
0

Homework Statement


An asteroid is moving along a straight line. A force acts along the displacement of the asteroid and slows it down. The asteroid has a mass of 4.5*10^4 kg, and the force causes its speed to change from 7100m/s to 5500m/s.
(a) What is the work done by the force?
(b) If the asteroid slows down over a distance of 1.8*10^6m, Determine the magnitude of the force.

Homework Equations


W=F*cos(theta)*s
F=ma
instantaneous acceleration = delta(v)/delta(t) as limit of t approaches 0.
W=KEf-KEo= 1/2MVf^2 - 1/2MVo^2

The Attempt at a Solution


First, I looked for the instantaneous acceleration created by the force by subtracting 7100m/s - 5500m/s
a= 16,000 m/s. This means the force slowed down the asteroid by 1.6*10^4 m/s.
I multiply 45,000 kg by 16,000 m/s to calculate the force.
F= 7.2*10^8 N
I plug this into my work equation:
W=7.2*10^8*cos(180)*s

that's where I'm stuck because I'm not sure what my displacement is.
I tried the last equation and found initial Kinetic energy as Ke=1/2mv^2
so: 1/2(4.5*10^4kg)(7100m/s)^2
= 1.13*10^12
final kinetic is initial kinetic + work
1.13*10^12 + Fcos(theta)s

I feel like I may be over-complicating this question. I feel like I'm close but I'm missing something obvious. Help appreciated!
 
Physics news on Phys.org
Yes, you are over complicating things. You already have an expression for work in terms of the initial and final kinetic energies, what do you get if you use it? Once you have found the work, how can you compute the force from this?

Note: Never forget that physical quantities come with units, they are important and without them your results do not hold any meaning. ,ake it a habit to always write them out!
 
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...
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...
Back
Top