Video Transcript
A body of mass three kilograms is
moving under the action of a force 𝐅 such that its displacement 𝐬 of 𝑡 is equal
to five 𝑡 squared 𝐢 plus seven 𝑡 𝐣. Find the work done by this force in
the first six seconds of its motion, given that the displacement is measured in
meters, the force in newtons, and the time 𝑡 in seconds.
In this question, we are asked to
find the work done by a force. And we recall this can be
calculated by finding the scalar or dot product between the force vector 𝐅 and the
displacement vector 𝐝. In this question, the displacement
vector is 𝐬, and this is given in terms of time 𝑡. We are not given the value of the
force vector in the question. Recalling Newton′s second law, we
know that this force vector is equal to the mass of the body multiplied by its
acceleration vector. And the mass of the body is three
kilograms.
When given a displacement vector in
terms of time 𝑡, we can find a velocity vector 𝐯 of 𝑡 and acceleration vector 𝐚
of 𝑡 by differentiating. This is because the velocity 𝑣 is
equal to d𝑠 by d𝑡 and the acceleration 𝑎 is equal to d𝑣 by d𝑡, or d two 𝑠 by
d𝑡 squared. We begin with our expression for
displacement given in the question, five 𝑡 squared 𝐢 plus seven 𝑡 𝐣. Differentiating this term by term
with respect to 𝑡 gives us 10𝑡 𝐢 plus seven 𝐣. The velocity of the body 𝐯 of 𝑡
is equal to 10𝑡 𝐢 plus seven 𝐣. Differentiating once again, we see
that the acceleration of the body is simply equal to 10𝐢.
We are now in a position to
calculate the force vector 𝐅. This is equal to the mass of three
kilograms multiplied by the acceleration of 10𝐢 meters per second squared, which
gives us a force vector of 30𝐢. And this is measured in the
standard units of newtons. Recalling that we need to find the
work done by the force in the first six seconds, we need to calculate the
displacement of the body after six seconds. When 𝑡 is equal to zero, 𝐬 of 𝑡
is equal to zero 𝐢 plus zero 𝐣. This means that the body starts at
the origin. When 𝑡 is equal to six seconds,
the displacement is equal to five multiplied by six squared 𝐢 plus seven multiplied
by six 𝐣. This is equal to 180𝐢 plus
42𝐣.
Since the body started at the
origin, we now have the displacement vector in the first six seconds of motion. Since we also have the force
vector, we can find the scalar product of these to calculate the work done. The work done is equal to the
scalar product of 30𝐢 and 180𝐢 plus 42𝐣. To calculate the scalar product, we
find the sum of the products of the individual components. This is equal to 30 multiplied by
180 plus zero multiplied by 42, which is equal to 5400.
The work done by the force in the
first six seconds of motion is 5400 joules.
