A5kg mass at rest on the ground is rised up to 15m. find the work that was done on the object.

a. e before= work= e after=
b. conservation of energy equation:
c. solve.

E before = 0 J Work = 735.75 J E after= 735.75 J

Explanation:

The work that was done on the object is calculated as follows:

W = m*g*h

where m refers to the mass of the object, g to gravitational acceleration (9.81 m/(s^2)) and h to the hight that the object was raised up. Replacing with data into the equation (units are ommited):

W = 5*9.81*15

W = 735.75 J

Taking the ground as heigth = 0 m, then both potential and kinetic energy of the object before the lift was zero (notice that it was resting). Conservation energy equation states that the change of energy of an object is equal to the work applied on the object (like this case) or done by the object. That is:

ΔE = W

where ΔE = Energy after - Energy before

Therefore:

Energy after = W = 735.75 J

The potential energy after the lift is also calculated as m*g*h, so the energy after the lift is all potential energy gained by the object due to the work applied on it.

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west

explanation:

wheres the pic