Respuesta :
Answer:
181.11 m
Explanation:
Net Force is the sum of all the resultant force of all the forces acting on a body. The net force experienced by the jet is the difference of the Applied force and frictional force and this is represented in equation 1;
[tex]F_{net} = F_{applied} - F_{friction} \\[/tex] ............................1
but [tex]F_{friction}[/tex] = μmg .......................................2
Where μ is the coefficient of friction = 0.1
m is the mass of the body = 72 kg
g is the acceleration due to gravity 9.81 m/[tex]s^{2}[/tex]
Substituting into equation 2 we have;
[tex]F_{friction}[/tex] = 0.1 x 72 x 9.81
[tex]F_{friction}[/tex] = 70.63 N
Now we substitute our answer in equation 1;
[tex]F_{net} = 190 N - 70.63 N \\[/tex]
[tex]F_{net}[/tex] = 119.37 N
We have to calculate the net acceleration in order to get our velocity .
[tex]F_{net}[/tex] = m[tex]a_{net}[/tex]
[tex]a_{net}[/tex] = [tex]\frac{F_{net}}{m}[/tex]
[tex]a_{net}[/tex] = [tex]\frac{119.37 N}{72 kg }[/tex]
[tex]a_{net}[/tex] = 1.66 [tex]m/s^{2}[/tex]
Speed can be express as;
v = u + at ...................3
where u is the initial velocity, which is 0 in this case.
a is the net acceleration = 1.66 [tex]m/s^{2}[/tex]
t is the time which is 9 s
substituting the values in equation 3 we have
v = 1.66 [tex]m/s^{2}[/tex] x 9 s
v = 14.94 m/s
Calculating for Sam's distance for the first 9 seconds, using the equation of motion we have;
[tex]S_{1} = ut +\frac{1}{2}at^{2}[/tex]
the jet was initially at rest so initial velocity is 0 and [tex]a_{net}[/tex] = 1.66 [tex]m/s^{2}[/tex]
[tex]S_{1} = \frac{1}{2} *1.66 m/s^{2} *9^{2}[/tex]
[tex]S_{1}[/tex] = 67.23 m
Also we have to calculate Sam's distance after nine seconds using equations of motion express below;
[tex]v^{2} -u^{2} = 2as[/tex]
making S the subject formula we have;
[tex]S_{2} =\frac{v^{2}-u^{2} }{2a}[/tex] ....................................4
v is the maximum velocity after the fuel finished and its 14.94 m/s and a is the acceleration along the horizontal plane which put into consideration the coefficient of friction. a = μg = 0.1*9.8 m/[tex]s^{2}[/tex] = 0.98 m/[tex]s^{2}[/tex]
We substitute our values into equation 4 to get our remaining distance;
[tex]S_{2} = \frac{(14.94 m/s)^{2} }{2(0.98 m/s^{2} )}[/tex]
[tex]S_{2}[/tex] = 113.88 m
Therefore the total distance S = [tex]S_{1}[/tex] + [tex]S_{2}[/tex]
S = 67.23 m + 113.88 m
The total distance covered by Sam is 181.11 m
Given Information:
Mass = 72 kg
Force exerted by ski = 190 N
coefficient of kinetic friction = 0.1
ski runs out of fuel after = 9 sec
Required Information:
Distance traveled when he stops = ?
Answer:
d = 179.33 m
Explanation:
The problem can be divided into two parts
Part 1: before running out of fuel
In the first part, the forces acting on Sam are
Ski thrust and Force of friction
Force of friction = kmg
where k is the coefficient of kinetic friction, m is Sam's mass and g is gravity 9.8 m/s²
Force of friction = 0.1*72*9.8
Force of friction = 70.56 N
so the net force acting on Sam is
Fnet = ski thrust - force of friction
Fnet = 190 - 70.56
Fnet = 119.44 N
According to Newton's second law of motion
F = ma
a = F/m
a = 119.44/72
a = 1.65 m/s²
The distance traveled can be found using kinematic equation
d = vi + 0.5at²
d = 0 + 0.5*1.65*9²
d = 66.82 m
The speed at this point is
vf = (d + 0.5at²)/t
vf = (66.82 + 0.5(1.65)(9)²)/9
vf = 14.85 m/s
Part 2: after running out of fuel
When the fuel runs out the ski is no longer applying any force so the only force acting on Sam is force of friction
kmg = ma
kg = a
a = 0.1*-9.8
a = -0.98 (minus sign due to deceleration)
The distance traveled can be found using kinematic equation
2ad = vf² - vi²
d = (vf² - vi²)/2a
d = (vf² - vi²)/2a
d = (0² - 14.85²)/2*-0.98
d = 112.51 m
How far has Sam traveled when he finally coasts to a stop?
We have to sum the distance before and after Sam runs out of fuel because he stops after covering the sum of these distance
d = 66.82 + 112.51
d = 179.33 m
