An Asteroid is moving along a straight line

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SUMMARY

An asteroid with a mass of 4.5 x 104 kg experiences a force that reduces its speed from 7100 m/s to 5500 m/s over a distance of 1.8 x 106 m. The work done by the force can be calculated using the kinetic energy equations, specifically W = KEf - KEo. The calculated force acting on the asteroid is 7.2 x 108 N, derived from the change in velocity and mass. Proper unit notation is crucial for accurate results in physics calculations.

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Familiarity with kinetic energy equations (KE = 1/2 mv2)
  • Knowledge of work-energy principle (W = KEf - KEo)
  • Ability to perform calculations involving significant figures and units
NEXT STEPS
  • Review the work-energy theorem in classical mechanics
  • Learn about the implications of force and displacement in physics
  • Study the concept of instantaneous acceleration and its calculation
  • Explore the importance of unit consistency in physical equations
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of force and work calculations in real-world scenarios.

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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!
 

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