Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Experience the convenience of finding accurate answers to your questions from knowledgeable professionals on our platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
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 our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.
