Video Transcript
A block moves in a straight line
under the action of a force 𝐅 equal to 12𝑠 squared plus six 𝑠 plus 𝑐 newtons,
where 𝑠 meters is the displacement of the body from its initial position. The work done by the force in
moving the block from 𝑠 equals zero meters to 𝑠 equals three meters is 34
joules. Determine the work done by 𝐅 in
moving the block from 𝑠 equals three meters to 𝑠 equals six meters.
As we have a force acting on an
object moving in a straight line, we can use the formula 𝑊 is equal to the integral
of 𝐅 with respect to 𝑠 to calculate the work done by the force. In this question, we are told that
the force 𝐅 is equal to 12𝑠 squared plus six 𝑠 plus 𝑐 newtons. The work done is therefore equal to
the integral of this with respect to 𝑠. Integrating each term with respect
to 𝑠, we have four 𝑠 cubed plus three 𝑠 squared plus 𝑐𝑠. We are told that the work done by
the force in moving the block from 𝑠 equals zero meters to 𝑠 equals three meters
is 34 joules. We can therefore substitute in
these values as shown as this will enable us to calculate the constant 𝑐.
When 𝑠 equals three, the
right-hand side of our equation becomes 135 plus three 𝑐. And when 𝑠 equals zero, this
expression equals zero. This means that 34 is equal to 135
plus three 𝑐. Subtracting 135 from both sides of
this equation, we have three 𝑐 is equal to negative 101. We can then divide through by three
such that 𝑐 is equal to negative 101 over three. Substituting this back in to the
expression for the work done, we have 𝑊 is equal to four 𝑠 cubed plus three 𝑠
squared minus 101 over three 𝑠.
As we need to calculate the work
done by 𝐅 from 𝑠 equals three to 𝑠 equals six, we can substitute these values
into our expression. When 𝑠 equals six 𝑊 is equal to
770, and when 𝑠 equals three, 𝑊 is equal to 34. The work done from 𝑠 equals three
meters to 𝑠 equals six meters is therefore equal to 770 minus 34, which is equal to
736. The final answer is equal to 736
joules.
