Question

The train,of 2.5 kg is stopped by compressing a spring in the buffer.after the train has stopped, the energy stored in the spring is 0.48J

Answer 1

$</p><p></p><p></p><p> </p><p>PE=KE</p><p> PE = \frac{1}{2}mv2 PE= </p><p>2</p><p>1</p><p> mv </p><p>2</p><p> </p><p> v = \sqrt{\frac{2 × PE}{m}} v= </p><p>m</p><p>2×PE</p><p></p><p>$

v=

2.5

2×0.48

v = 0.384 v=0.384

Hence, initial speed of the train v = 0.384 v=0.384 m/s