Question

1. A 250g bullet is fired with a speed of 300m/s. if it stopped after hitting a target 150m away, (a) what is its initial kinetic energy? (b) What is the magnitude of the force that stopped it? 2. Determine the GPE of a 0.5kg object when it is raised (a.) 0.75m from the floor. (b) What happens to its GPE if we double the height? 3. How high could a 40kg mass be lifted if an energy of 2x10^6 J is used in lifting it?

Answer 1

Answer:

See below

Explanation:

KE = 1/2 m v ^2

    = 1/2 (250/1000) 300^2   j         the bullet is 250/1000  kg  ( assuming g is grams and not GRAINS - in which bullet weight is usually measured)

force to stop it is the same number

PE = mgh = .5 *9.81  *  .75 =

     doubling height will double PE

2 x 10^6  =  40 * 9.81  *  h      sove for 'h' in meters  

Answer 2

Initial kinetic energy is 11250 J and magnitude of force to stop bullet is 75 N.

The GPE for 0.75m height floor is 3.75J.

The GPE for double of 0.75m height is 7.5J.

The high on which 40 kg mass is lifted is 5000m.

What is kinetic energy?

The kinetic energy of an object is the energy that it possesses due to its motion.

$KE = 1/2 m v ^2$

      $=1/2 (250/1000) 300^2 J\\= 11250 J$

Work done,

$W= fs\\\\f=\frac{W}{s}\\\\f=\frac{11250}{150}$

$f=75 N$

What is GPE?

Gravitational potential energy is the potential energy a massive object has in relation to another massive object due to gravity.

PE = mgh

    = .5 *9.81  *  .75

    = 3.75 J

Potential energy when height doubles = 7.5J

What is height to which object lift?

The potential energy is given by

$PE = mgh\\\\h = PE/ mg\\\\h = 2*10^6 J /40*9.8\\\\h = 5000 m$

To learn more about Kinetic and potential energy here

https://brainly.com/question/15764612

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