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