Answer:
a ) = 381.48 J
b )= 84.25 cm
Explanation:
Kinetic energy of the runner
= 1/2 m v²
= .5 x 66 x 3.4²
= 381.48 J
The final kinetic energy of the runner is zero .
Loss of mechanical energy
= 381.48 J
This loss in mechanical energy is due to action of frictional force .
b )
Let s be the distance of slide
deceleration due to frictional force
= μmg/m
.7 x 66 x 9.8 / 66
a = - 6.86 m s⁻¹
v² = u² - 2 a s
0 = 3.4² - 2x6.86 s
s = 3.4² / 2x6.86
= .8425 m
84.25 cm
Since the ladder is standing, we know that the coefficient
of friction is at least something. This [gotta be at least this] friction
coefficient can be calculated. As the man begins to climb the ladder, the
friction can even be less than the free-standing friction coefficient. However,
as the man climbs the ladder, more and more friction is required. Since he
eventually slips, we know that friction is less than what's required at the top
of the ladder.
The only "answer" to this problem is putting lower
and upper bounds on the coefficient. For the lower one, find how much friction
the ladder needs to stand by itself. For the most that friction could be, find
what friction is when the man reaches the top of the ladder.
Ff = uN1
Fx = 0 = Ff + N2
Fy = 0 = N1 – 400 – 864
N1 = 1264 N
Torque balance
T = 0 = N2(12)sin(60) – 400(6)cos(60) – 864(7.8)cos(60)
N2 = 439 N
Ff = 439= u N1
U = 440 / 1264 = 0.3481
Answer:
ΔR = 9 s
Explanation:
To calculate the propagation of the uncertainty or absolute error, the variation with each parameter must be calculated and the but of the cases must be found, which is done by taking the absolute value
The given expression is R = 2A / B
the uncertainty is ΔR = | 
| ΔA + | 
| ΔB
we look for the derivatives
= 9 / B
= 9A ( 
)
we substitute
ΔR = 
ΔA + 
ΔB
the values are
ΔA = 2 s
ΔB = 3 s
ΔR = 
2 + 
3
ΔR = 1.636 + 7.14
ΔR = 8,776 s
the absolute error must be given with a significant figure
ΔR = 9 s
13/2 = 6.5 Km/h
That’s 6500/3600= 1.8 m/s
