I need an answer for this plzz!!number 2 anybody can help ??

1 answer
Question:

I need an answer for this plzz!!
number 2
anybody can help ??

Answers

2.1) (i) W = mg downwards (ii) N = R = Normal Reaction from the ground upwards (iii) Fe = Force of engine towards the right (iv) f = friction towards the left (v) ma = Constant acceleration towards right.2.2.1)v = 25 m/su = 0 m/s∆v = v - u = (25 - 0) m/s = 25 m/sx = X∆t = 50 s[tex]a \: = \: \frac{dv}{dt} \: = \: \frac{25 \: \frac{m}{s} }{50 \: s} \: = \: 0.5 \: \frac{m}{ {s}^{2} } [/tex]a = 0.5 m/s².2.2.2)F = ma = 900 kg × 0.5 m/s² = 450 N.2.2.3)[tex]2ax \: = \: {v}^{2} \: - \: {u}^{2} [/tex][tex]x \: = \: \frac{ {v}^{2} \: - \: {u}^{2} }{2a} \: = \: \frac{{(25 \: \frac{m}{s})}^{2} \: - \: {(0 \: \frac{m}{s} )}^{2} }{2 \: \times \: 0.5 \: \frac{m}{ {s}^{2} } } \: = \: 625 \: m[/tex]2.3)Fe = f + maFe - f = maFor velocity to be constant,a should be 0, or, a = 0,Fe = f = 270 N2.4.1)v = 0u = 25 m/sa = -0.5 m/s²v = u + att = -u/a = -(25)/(-0.5) = 50 s.2.4.2)x = -625/(2×(-0.5)) = 625 m.

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