Question

Sam, whose mass is 75kg , takes off across level snow on his jet-powered skis. The skis have a thrust of 220N and a coefficient of kinetic friction on snow of 0.1. Unfortunately, the skis run out of fuel after only 11s .

a) what is Sam's top speed?

b) How far has Sam traveled when he finally coasts to a stop?

Answer 1

for the first ten seconds, the net force acting on sam is
220N - force of friction

the force of friciton = u mg where u is the coefficient of friction, so the net force is

220- 0.1x75kgx9.8m/s/s = 220N- 73.5N=146.5N

therefore his accelerationin the first ten secs is

a = F/m=146.5N/75kg=1.9533m/s/s

so after 10 secs he reaches a top speed of 1.9533m/s/s x 11s = 21.4866m/s

in this ten sec interval he travels a distance of 1/2 at^2 =
1/2(1.9533)11^2=118.17465m

when his fuel runs out, the only force acting on him is the force of friction which is u mg

so his acceleration is now

u mg = ma or a=ug =-0.98 m/s/s (the minus sign means he is slowing down)

the distance he will travel during this phase is given by

vf^2=v0^2+2ad

vf=final speed=0
v0=initial speed = 21.4866 m/s since this is the speed he acquired just before his fuel cut off
a=acceleraion =-0.98m/s/s
d=distance traveled in this phase

we have

0=21.4866^2+2(-0.98)d=> d=21.4866^2/(2*0.98) = 235.547m

his total distance traveled = 118.17465m+235.547m = 353.7225m