Answer
given,
time taken to stop by the object = 5 s
distance travel before stopping = 60 m
final velocity = 0
using equation of motion
v = u + at
0 = u - 5 a
[tex]a_x = \dfrac{u}{5}[/tex]
[tex]s = u t + \dfrac{1}{2}at^2[/tex]
[tex]s = 5u + \dfrac{1}{2}\times \dfrac{4}{5}\times u^2[/tex]
[tex]60 = 2.5 u[/tex]
u = 24 m/s
[tex]a_x = \dfrac{24}{5}[/tex]
a_x = 4.8 m/s²
b) using energy conservation
[tex]\dfrac{1}{2}mv_i^2 + \dfrac{1}{2}mv_f^2 = mg(\Delta h)[/tex]
[tex]\dfrac{1}{2}v_i^2= g(\Delta h)[/tex]
[tex](\Delta h) = \dfrac{0.5 \times 24^2}{9.8}[/tex]
Δh = 29.38 m
(a) The speed of the object at the bottom of the incline is 24 m/s and its acceleration along the horizontal surface is 4.8 m/s².
(b) The height of the incline is 29.4 m.
The speed of the object at the bottom of the ramp is calculated as follows;
v = u + at ---(1)
where;
v = at
v = 5a
Second equation
v² = u² + 2as
v² = 0 + 2as
v² = 2as
(5a)² = 2a(60)
25a² = 120a
25a = 120
a = 120/25
a = 4.8 m/s²
v = 5 x 4.8
v = 24 m/s
The height of the incline is determined from the conservation of energy,
mgh = ¹/₂mv²
gh = ¹/₂v²
h = (v²)/(2g)
h = (24²)/(2 x 9.8)
h = 29.4 m
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