Question

you drop a 2 kg book to a friend who stand on the ground at distance D=10.0 m below, if your friends out streched hand at distance d=1.50 m above the ground. (a) How does the work Wg does the gravitational force do on the book as it is drop to her hands? (b) what is the change U in the gravitational potential energy of book earth system drop? if the gravitational potential energy U of that system is taken be zero by ground level

Answer 1

Work is defined as force*displacement so total d=8.5m and force will be 2kg*g which is roughly 20N of force so W = 20N*8.5m=170J

For part b the gravitational energy is m*g*h so gravitational energy = 200J since 2kg*10 m/s^2 *10m so the change since the work is 170 j means total change is 170J and 30J is left

Answer 2

Part(a) The work done by gravitational force on the book as it is dropped to her hands is 167 Joules

Part(b) The change in the gravitational potential energy(
`\Delta U`) of book earth system is -167 Joule

Step-by-step explanation:

Part (a)

Since the work done (W)= Force(F)x Displacement(D) in the direction of force

Here force is applied by gravity in downward direction (towards the center of the earth),
`F_g=mg`

And displacement in the direction of force is D= (10-1.5) m = 8.5 m

Therefore work done ,
`W_g=F_gD=mgD`

where mass, m =2 kg and acceleration due to earth's gravity ,
`g=9.81(m)/(s^(2))`

`\therefore W_g=mgD=2* 9.81* 8.5 Joules`

=>
`W_g=167 Joules`

Thus the work done by gravitational force on the book as it is dropped to her hands is 167 Joules

Part (b)

By the work energy theorem , Change in the gravitational potential energy (
`\Delta U`)= - Work done by gravitational force (
`W_g`)

=>
`\Delta U=-W_g=-167 Joules`

Thus the change in the gravitational potential energy(
`\Delta U`) of book earth system is -167 Joule