Answer:
Part(i) the time taken for this cart to reach the bottom of the inclined plane is 1.457 s
Part(ii) the velocity of the cart when it reaches the bottom of the inclined plane is 4.531 m/s
Part(iii) the kinetic energy of the cart when it reaches the bottom of the inclined plane is 25.663 J
Explanation:
Given;
mass of the cart = 2.5 kg
angle of inclination, β = 18.5⁰
length of inclined plane = 3.3m
Part(i) the time taken for this cart to reach the bottom of the inclined plane
s = ut + ¹/₂×at²
initial vertical velocity, u = 0
s = 3.3 m
s = ¹/₂×at²
[tex]t = \sqrt{\frac{2s}{a} }[/tex]
acceleration, of the cart, a = gsinβ
a = 9.8sin(18.5) = 3.11 m/s²
[tex]t = \sqrt{\frac{2X3.3}{3.11 }}= 1.457 s[/tex]
Part(ii) the velocity of the cart when it reaches the bottom of the inclined plane
V = a×t
V = 3.11 × 1.457 = 4.531 m/s
Part(iii) the kinetic energy of the cart when it reaches the bottom of the inclined plane
KE = ¹/₂MV²
= ¹/₂ × 2.5× (4.531)²
= 25.663 J
