Answer:
a) 47.63 m
b) 188572.725 joule
c) 373.88 N
Explanation:
Given data:
initial speed = 110 km/hr = 110,000 m/3600 s = 30.6 m/s
[tex]E_{inital} = E_{final}[/tex]
[tex]mgy_i + \frac{1}{2} mv_i^2 = mgy_f + \frac{1}{2} mv_f^2[/tex]
solving y_f we have
[tex]y_f = \frac{m(gy_i +\frac{1}{2} v_i^2 - \frac{1}{2} v_f^2}{mg}[/tex]
[tex]y_f = \frac{(0 + \frac{1}{2}(110 \times \frac{1000}{60\times 60})^2 - 0}{9.8})[/tex]
[tex]y_f = h = 47.63 m[/tex]
b) loss of energy = m g (47.63 - 22)
= 750(9.81)(25.63) = 188572.725 Joules lost
c) work done = force acted in motion direction × distance moved
[tex]188572.725 = F (\frac{22}{sin2.5})[/tex]
F = 373.88 N
