At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Get expert answers to your questions quickly and accurately from our dedicated community of professionals. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
Complete question is;
A 5 × 10^(5) kg subway train is brought to a stop from a speed of 0.5 m/s in 0.4 m by a large spring bumper at the end of its track. what is the force constant k of the spring
Answer:
781250 N/m
Explanation:
From conservation of energy, potential energy is equal to kinetic energy.
Thus;
½mv² = ½kx²
where;
m = mass of train
v = velocity of train
k = force constant of spring
x = the distance the train went while being stopped
We are given;
Mass; m = 5 × 10^(5) kg
Velocity; v = 0.5 m/s
Distance; x = 0.4 m
Thus, from ½mv² = ½kx²
Divide both sides by ½ to get;
mv² = kx²
k = mv²/x²
k = [(5 × 10^(5)) × 0.5²]/0.4²
k = 781250 N/m
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
