Respuesta :
Quite a few things can be determined using this information.
(i) First, we can calculate the Weight of the fireman using W = mg.
We get its magnitude as W = (80)(9.8) = 784 N
(ii) While his Weight is responsible for pulling him down, the pole exerts a constant Resistive Force that is given to be 500 N.
We can calculate the Net Force acting on the fireman as
[tex]F_{net} = W - Resistive Force[/tex]
We get its numerical value to be [tex]F_{net} = 284N[/tex]
(iii) Using this, we can calculate the acceleration with which he slides down the pole from Newton's 2nd law equation as
[tex]F_{net} =ma[/tex]
Therefore, [tex]a = \frac{284}{80} =3.55 m/s^{2}[/tex]
(iv) With this acceleration, he slides down a distance of 5.0 m - 0.4 m = 4.6 m before he starts applying an additional force with his knees.
We can calculate the velocity he attains just before bending his knees using the following data:
Initial Velocity at the top of the pole [tex]V_{i} =0[/tex]
Vertical displacement down the pole D = 4.6 m
Acceleration [tex]a = 3.55 m/s^{2}[/tex]
Final Velocity [tex]V_{f} = ?[/tex]
Using the equation [tex]V^{2} _{f} =V^{2} _{i} +2aD[/tex]
Plugging in the numbers, we have [tex]V^{2} _{f} =0+2(3.55)(4.6)[/tex]
Thus, we get the value of [tex]V_{f}[/tex] as 5.72 m/s
(v) This velocity serves as the initial velocity for the part of the journey with his knees bent.
We can calculate the acceleration he has using the following data:
Initial Velocity [tex]V_{i}=5.72[/tex] m/s
Final Velocity [tex]V_{f} =0[/tex]
Displacement during the last part of the journey D = 0.4 m
Acceleration a = ?
Again using the equation [tex]V^{2} _{f} =V^{2} _{i} +2aD[/tex], and plugging in the known numbers, we get
[tex](0)^{2} =(5.72)^{2} +2(a)(0.4)[/tex]
Hence, [tex]a=-40.898 m/s^{2}[/tex]
(vi) We can calculate the Net Force acting on him during this part of the journey as
[tex]F_{net} =(80)(-40.898)=-3271.84N[/tex]
(vii) Since the Net Force is the vector sum of Weight, Resistive Force, and the additional Knee Force, we can write them as
[tex]W-Resistive Force-Knee Force=-3271.84N[/tex]
Solving for Knee Force gives its magnitude to be 3555.84 N.
Thus, the fireman begins applying an additional 3,556 N force to stop himself in 0.4 m.
(viii) We can even calculate the time taken for the entire journey. We will deal with it in two parts.
For part one, the following information can be used:
Initial Velocity [tex]V_{i} =0[/tex]
Final Velocity [tex]V_{f} =5.72m/s[/tex]
Acceleration [tex]a = 3.55m/s^{2}[/tex]
Time taken t = ?
Using the equation [tex]V_{f} =V_{i} +at[/tex], we get the time taken as 1.6 seconds.
For the second part of the journey, the following information can be used:
Initial Velocity [tex]V_{i} =5.72m/s[/tex]
Final Velocity [tex]V_{f} =0[/tex]
Acceleration [tex]a=-40.898m/s^{2}[/tex]
Time taken t = ?
Using the equation [tex]V_{f} =V_{i} +at[/tex] again, and plugging in the appropriate values, we get the time taken as 0.14 seconds.
Hence, the total time the fireman took to slide down the 5.0 m pole is 1.74 seconds.
Hope this helps!
