a) as soon as you read "net force", you should consider the vector addition of forces. since the cyclist is moving down the incline, there is no motion perpendicular to the incline, hence the net force in the perpendicular direction is zero. the net force in the parallel-to-incline direction is the vector sum of the parallel component of the cyclist's weight (Wp) and the friction (F). The weight of the cyclist is the force that the earth exerts on him due to gravity, W=mg=100g where g is the acceleration due to gravity (10 ms-2). to find Wp, we need to use trigonometry (nearly every inclined plane question needs this).
now, Wp= 100gsin(20 deg). drawing a labelled force diagram and figuring out the angles will help u understand where this comes from.
hence Fnet= Wp-F
=100*10*sin(20 deg)-300
= 42 N
b) Here, our answer to part a helps. recall that Fnet=ma where m is the mass of the object and a is the acceleration. we know m and Fnet, so we can transpose for a: a=Fnet/m.
hence a=42/100
=0.42 ms-2
c) whenever a distance or time taken to move along an incline comes up, think "constant acceleration formulae" this is what we need to use here. we know:
u= 0 ms-1
t= 15 s
a= 0.42 ms-2
v= ?
to figure out which one of the five to use, always look at what we don't need at all. here, we neither know nor need to know the distance travelled (s or x), hence use the formula without s:
v=u+at
sub in:
v= 0+0.42*12
= 5 m/s
conceptually, the cyclist starts from rest and gains 0.42 metres per second in speed every second. hence after 12 seconds its speed will be 0.42*12.
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