W = _|....F*dx*cos(a)........With F=force, x=distance over which force acts on object,
.......0.............................and a=angle between force and direction of travel.
Since the force is constant in this case we don't need the equation to be an integral expression, and since the force in question - the force of friction - is always precisely opposite the direction of travel (which makes (a) equal to 180 deg, and cos(a) equal to -1) the equation can be rewritted like so:
W = F*x*(-1) ............ or ............. W = -F*x
The force of friction is given by the equation: Ffriction = Fnormal*(coeff of friction)
Also, note that the total work is the sum of all 45 passes by the sandpaper. So our final equation, when Ffriction is substituted, is:
W = (-45)(Fnormal)(coeff of friction)(distance)
W = (-45)...(1.8N).........(0.92).........(0.15m)
W = ................-11.178 Joules
The fraction of energy that is lost is 25%, it depends how fast the ball was going until it lost 25% of its energy, the gravitational energy was transferred into the kinetic energy that helped the ball bounce back
Answer:
The unrealistically large acceleration experienced by the space travelers during their launch is 2.7 x 10⁵ m/s².
How many times stronger than gravity is this force? 2.79 x 10⁴ g.
Explanation:
given information:
s = 220 m
final speed, vf = 10.97 km/s = 10970 m/s
g = 9.8 m/s²
he unrealistically large acceleration experienced by the space travelers during their launch
vf² = v₀²+2as, v₀ = 0
vf² = 2as
a =vf²/2s
= (10970)²/(2x220)
= 2.7 x 10⁵ m/s²
Compare your answer with the free-fall acceleration
a/g = 2.7 x 10⁵/9.8
a/g = 2.79 x 10⁴
a = 2.79 x 10⁴ g
