# A 5.0-kg bowling ball is moving at 6.0 m/s. In order to change its speed to 10.0 m/s, the net work done on the?

A 5.0-kg bowling ball is moving at 6.0 m/s. In order to change its speed to 10.0 m/s, the net work done on the bowling ball must be:

A. 40 J

B. 90 J

C. 160 J

D. 400 J

E. 550 J

F. None of the above

A bullet with a mass of 3.0 gram, moving at speed of 300 m/s, hits and embeds itself in the trunk of a tree. The tree exerts an average force of 250 N on the bullet. What is the distance the bullet travels inside the trunk?

A. 0.27 m

B. 0.54 m

C. 0.81 m

D. 1.1 m

E. 1.3 m

F. None of the above

A 5 kg projectile is fired at an angle of 25º above the horizontal. Its initial velocity is 200 m/s and just before it hits the ground its velocity is 150 m/s. What is the change in the internal energy of the projectile?

A. +19,000 J

B. -19,000 J

C. +44,000 J

D. -44,000 J

E. 0 J

56 J of work was done to increase the kinetic energy of a 1.90-kg mass. If the object's initial velocity was zero, what is its speed now?

A. 59 m/s

B. 56 m/s

C. 1.9 m/s

D. 7.7 m/s

E. None of the above

### 1 Answer

- Anonymous9 years agoFavorite Answer
First off, it's apparent that you are posting a good portion or all of your homework set to have someone else do it for you. I'll happily answer your questions, but you're never going to learn this material (or be able to do it comfortably on a test) if you don't practice it yourself. Homework points are small, test points are big.

1.) C. 160 J

The key to this problem is the work-kinetic energy theorem, which states W = delta(KE), meaning, the amount of work done is equal to the change in kinetic energy. Since KE = 0.5 mV^2, you can relate the masses and velocities to the work.

2.) B. 0.54 m

Again, this is the work-kinetic energy theorem, except now you are breaking up the left hand side of the equation up too, so that W = F*distance. You can then find the distance the projectile travels into the tree by F*distance = 0.5 * m*(V2^2 - V1^2)

3.) D. -44000 J

The total internal energy is the sum of the kinetic and potential energy. Since the potential energy doesn't change from state one to state two, you only need the equation: KE1 = KE2, and then:

delta(KE) = KE2-KE1 = 0.5 *m*(V2^2 - V1^2). Note, that you are losing energy from state one to two, so the sign will be negative (because V1 > V2).

4.) A. 58.9 m/s

The key to this problem is once again, the work kinetic energy theorem. Simple solve: W = 0.5 * m * (V2^2 - V1^2) for V2.