Question

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

Answer 1

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/s
u = 0 m/s
∆v = v - u = (25 - 0) m/s = 25 m/s
x = X
∆t = 50 s

`a \: = \: (dv)/(dt) \: = \: (25 \: (m)/(s) )/(50 \: s) \: = \: 0.5 \: \frac{m}{ {s}^(2) }`
a = 0.5 m/s².
2.2.2)
F = ma = 900 kg × 0.5 m/s² = 450 N.
2.2.3)

`2ax \: = \: {v}^(2) \: - \: {u}^(2)`

`x \: = \: \frac{ {v}^(2) \: - \: {u}^(2) }{2a} \: = \: \frac{{(25 \: (m)/(s))}^(2) \: - \: {(0 \: (m)/(s) )}^(2) }{2 \: * \: 0.5 \: \frac{m}{ {s}^(2) } } \: = \: 625 \: m`
2.3)
Fe = f + ma
Fe - f = ma
For velocity to be constant,
a should be 0, or, a = 0,
Fe = f = 270 N
2.4.1)
v = 0
u = 25 m/s
a = -0.5 m/s²
v = u + at
t = -u/a = -(25)/(-0.5) = 50 s.
2.4.2)
x = -625/(2×(-0.5)) = 625 m.