To solve this problem, it is necessary to apply the concepts related to the work done by a body when a certain distance is displaced and the conservation of energy when it is consumed in kinetic and potential energy mode in the final and initial state. The energy conservation equation is given by:
[tex]\Delta KE_i + \Delta PE_i = \Delta KE_f + \Delta PE_f[/tex]
Where,
KE = Kinetic Energy (Initial and Final)
PE = Potential Energy (Initial and Final)
And the other hand we have the Work energy theorem given by
[tex]\Delta KE = W = F*d[/tex]
Where
W= Work
F = Force
D = displacement,
PART A) Using the conservation of momentum we can find the speed, so
[tex]\Delta KE_i + \Delta PE_i = \Delta KE_f + \Delta PE_f[/tex]
[tex]\frac{1}{2}mv_1^2 + mgh_i = \frac{1}{2}mv_f^2+mg_h2[/tex]
The height at the end is 0m. Then replacing our values
[tex]\frac{1}{2}mv_1^2 + mgh_i = \frac{1}{2}mv_f^2+0[/tex]
Deleting the mass in both sides,
[tex]\frac{1}{2}v_1^2 + gh_i = \frac{1}{2}v_f^2[/tex]
Re-arrange for find [tex]v_f^2,[/tex]
[tex]v_f^2= 2(gh_i)+v_1^2[/tex]
[tex]v_f^2 = 2(9.8*165)+(175)^2[/tex]
[tex]v_f=\sqrt{33859}[/tex]
[tex]v_f = 184.008m/s[/tex]
PART B) Applying the previous Energy Theorem,
[tex]\Delta KE = W = F*d[/tex]
[tex]\frac{1}{2}mv^2 = F*d[/tex]
[tex]\frac{1}{2}(0.2)(184.008)^2 = (75)*d[/tex]
Solving for d
[tex]d = 45.15 m[/tex]
