Question

A bowling ball weighing 10.0 N is dropped off the top of a 5.00 m tall building. After falling 1.00m, the book’s kinetic energy is?

a. 10.0 J b. 5.00 J c. 40.0 J d. 50.0 J

Answer 1

The ball's kinetic energy is a. 10 J

Step-by-step explanation:

Conservation of Mechanical Energy

In an isolated system is subject only to conservative forces, then the mechanical energy is constant.

The mechanical energy is calculated as:

E=K+U

Where K is the kinetic energy and U is the gravitational potential energy.

The gravitational potential energy of the body of mass m can be calculated by:

U=m.g.h

Recall m.g is the weight of the body, thus:

U=W.h

Where h is the height of the object.

The kinetic energy is calculated as:

`\displaystyle K=(1)/(2)m.v^2`

Where v is the speed of the object.

Initially, the bowling ball is at rest at a height of 5 m. Its gravitational potential energy is:

U1=10*5=50\ J

The kinetic energy is zero since v=0.

Thus the initial mechanical energy is:

E=50 J + 0 J = 50 J

When the object falls 1 m, its height is h=4 m. The new gravitational potential energy is:

U2=10*4=40 J

Since the mechanical energy is constant:

U2 + K2 = 50 J

Solving for K2:

K2 = 50 J - 40 J

K2 = 10 J

The ball's kinetic energy is a. 10 J