Question

A 0.3 Kg ball is initially at rest at the top left side of a frictionless incline plane: The incline is 0.25 m height and has a length of 0.65 m.

a) What is the initial Gravitational Potential Energy of the Ball at the top of the incline? b) What will be the final Kinetic energy of the ball at the bottom right side of the incline? c) If there were friction between the ball and the incline, what would happen to the final Kinetic energy of the ball at the bottom of the incline? Increase, decrease, or states the same?

Answer 1

a.)

Gravitational Potential Energy is given by,

PE = m*g*h

for ball given, m = mass of ball = 0.3 kg

g = 9.81 m/s^2

h = height of ball from ground = 0.25 m

So, initial Gravitational Potential Energy of ball(PEi) = 0.3*9.81*0.25 = 0.736

PEi = 0.74 J

b.)

From energy conservation,

PEi + KEi + W = PEf + KEf

here, PEi = initial Gravitational Potential Energy of ball = 0.74 J

KEi = initial kinetic Energy of ball = 0 (given, at rest)

KEf = final kinetic Energy of ball = ??

PEf = final Gravitational potential energy = 0 (at bottom, h = 0)

W = Work done on ball by any force = 0

So, 0.74 + 0 + 0 = 0 + KEf

final Kinetic energy of the ball at the bottom = KEf = 0.74 J

c.)

Now friction force will act on ball in opposite direction of motion of ball.

So, W = negative

then, from above energy conservation:

KEf = 0.74 - W

therefore, final Kinetic energy of the ball at the bottom of the incline will decrease.

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