Question

A ball of mass 5.0kg is lifted off the floor a distance of 1.7m. 1. What is the change in the gravitational potential energy of the ball? 2. Now you release the ball from rest, and it falls to the floor. What is the kinetic energy of the ball just before it hits the floor? 3. What is the velocity of ball just before it hits the floor?

Answer 1

Answer:

Explanation:

Change in gravitational energy of the ball = mgh

5 mutiply 10 multiply 1.7 = 85J

Potential energy at height = Kinetic energy at bottom

KE= 85J

Velocity

v=5.83m/s

Answer 2

• The change in the gravitational potential energy of the ball is 83.3Joules

• The kinetic energy of the ball just before it hits the floor is 83.3Joules

• The velocity of the ball just before it hits the floor is 5.77m/s

The formula for calculating the gravitational potential energy is expressed as:

$GPE =mgh$

m is the mass of the ball

g is the acceleration due to gravity

h is the height attained by the ball

a) Given the following

mass m = 5.0kg

g = 9.8m/s²

h = 1.7m

Substituting the given parameters into the formula

$GPE =5.0\times 9.8 \times 1.7\\GPE=83.3Joules$

Hence the change in the gravitational potential energy of the ball is 83.3Joules

b) To get the  kinetic energy of the ball just before it hits the floor, we must know that;

Potential energy at height = Kinetic energy at bottom = 83.3Joules

c) The formula for calculating the kinetic energy is expressed as:

$K.E=\frac{1}{2}mv^2\\K.E =\frac{1}{2}\times 5\timesv^2\\83.3 \times 2 = 5v^2\\ 166.6 = 5v^2\\v^2=33.32\\v=5.77m/s$

Hence the velocity of the ball just before it hits the floor is 5.77m/s

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