Question

On an essentially frictionless, horizontal ice rink, a skater moving at 6.0 m/s encounters a rough patch that reduces her speed by 46 % due to a friction force that is 26 % of her weight.

Use the work-energy theorem to find the length of this rough patch.

Answer 1

of 6 = 2.76 m/s, final speed-6-2.76 Solution:-u-6 m/s, 46% According to work energy theorem Initial energy Final energy +lost of energy Initial energy =-mu, =-m (6)2-18 m -3.24m/s Final energy-_ mu--× m (3.24)2-5.25m Energy lost in the friction-0.26x m × 9.8 x x Hence 18m = 5.25 m + 2.55 mx 18 = 5.25 + 2.55 x x 5 meter

Answer 2

v=6m/s, V_f= (.54)6= 3.24m/s 

IE=FE+LE

IE= 1/2mv²

FE= 1/2mV_f²

LE=friction multiplied delta x

delta x= length of rough patch = dx

friction=26%wieght= .26mg

1/2m6²=1/2m3.24²+.26mgdx

mass cancels 

1/2(6)²=1/2(3.24)²+.26(9.8)dx

dx=(18-5.25)/(.26(9.8))=5