Question

Sam, whose mass is 70kg , takes off across level snow on his jet-powered skis. The skis have a thrust of 280N and a coefficient of kinetic friction on snow of 0.15. Unfortunately, the skis run out of fuel after only 12s .How far has Sam traveled when he finally coasts to a stop?

Answer 1

forces along x- axis = F_T +f                           ( F_T = thrust, f = friction force)

F_T +f = ma                                          ( f = u*R = u*mg)

F_T + u*mg = ma

280+ 0.15* 70*(-9.8) = 70*a

a = 280- 0.15*70*9.8 / 70

a= 2.53 m/s²

distance travelled in 12 sec, s= ut + (1/2)*a*t²                         ( u= initial velocity=0)

s = 0+ (1/2)*2.53*12² = 182.16m

by V = u+at

V( top speed attained by Sam) = 0+ 2.53*12                          ( u= initial velocity=0)

V = 30.36m/s

B) now ,         ma= f                                          ( now their will be no upward thrust)

ma= u*mg

a = u*g = 0.15*(-9.8)

a= -1.47m/s²

By equation of motion,           V² - u² = 2*a*s

V(final velocity) = 0,              u( initial velocity) = 30.36 m/s

s'( distance travelled when fuel got finished) = 0- 30.36² / 2*(-1.47)

s' = 313.5 m

total distance travelled by sam = s+s' = 313.5 + 182.16

                                  = 495.6m