Question

A cart, which has a mass of m - 2.50 kg, is sitting at the top of an inclined plane which is 3.30 meters long and which meets the horizontal at an angle of B = 18.5
A. How long will it take for this cart to reach the bottom of the inclined plane?
B. What will be the velocity of the cart when it reaches the bottom of the incline?
C. What will be the kinetic energy of the cart when it reaches the bottom of the incline?

Answer 1

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

$t = \sqrt{\frac{2s}{a} }$

acceleration, of the cart, a = gsinβ

a = 9.8sin(18.5) = 3.11 m/s²

$t = \sqrt{\frac{2X3.3}{3.11 }}= 1.457 s$

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