Hello! A 17.8 kg curling stone is released at an initial speed of 1.95 m/s. The kinetic frictional force is 0.61 N. What will be the speed of the stone after in has traveled 38 m? A. 11.5 m/s B. 1.09 m/s C. 0.467 m/s D. Impossible physical situation This was an exam question from class. I got the answer correctly, but don\'t quite understand the calculation. I just know that the curling stone loses energy to the surrounding afterwards, so I guessed it. Thank you! Solution here, mass ,m = 17.8 kg intial speed , u = 1.95 m/s frictional force , ff = 0.61 N s = 38 m let the final speed be v work done by friction = change in kientic energy - F * s = 0.5 * m * ( v^2 - u^2) - 0.61 * 38 = 0.5 * 17.8 * ( v^2 - 1.95^2) solving for v v = 1.09 m/s the final speed is B) 1.09 m/s .