Question

A object of mass 50.0kg moves with a velocity of 11.2 m/s and an event occurs whereby it transfers 1,539 J of energy to another object as work .What is the new velocity of the object after the event.​

Answer 1

Answer:

28 v

Explanation:

Answer 2

The ending velocity of the object will be 7.99 m$s^{-1}$.

Explanation:

Given:

The beginning velocity of the object = 11.2 m$s^{-1}$

The mass = 50 kg

Energy transferred after the collision of the two objects= 1539 joules

To find = the newly calculated velocity of the object

Solution:

Let us first derive the energy of the moving object:

Kinetic energy = m×$v^{2}$ ÷2

The kinetic energy of the articles if given by= 50× $(11.2)^{2}$÷2

= 50×125.44÷2

6272÷2

Kinetic energy = 3136 joules

Thus, since we know the initial kinetic energy as well as the amount of energy transferred. We can simply find the resulting energy of the article.

Final energy = initial kinetic energy - the ending energy

Putting the values,

Final energy  = 3136-1539= 1597 joules

Thus, the speed of the object now can be calculated from the formula for kinetic energy derivation

Kinetic energy = m×$v^{2}$ ÷2

Putting the new values,

1597 = 50×$v^{2}$ ÷2

3194 = 50×$v^{2}$

$v^{2}$ = 63.88

v= 7.99 m$s^{-1}$

Result:

The final velocity of the article is calculated as 7.99 m$s^{-1}$.

(#SPJ2)