Question

4.
A block of mass 5 kg initially at rest at the
origin is acted upon by a force along the
positive X - direction represented by
F=(20 +5x)N. Calculate the work done by the
force during the displacement of the block
from x=0 to x=4m.​

Answer 1

Answer:

120 joule

Explanation:

we know smaller work done,dw=F.ds

as force and displacement are in same direction so angle is 0 degree

now dw= integration of Fds from x=0 to x=4

we get w= (20x+5xsquare/2) from x=0 to x=4

substituting value of x we get , W= 120 joule