Question

A particle of mass 0.01kg travels along a space curve with velocity given by $4\widehat i + 16\widehat k$m/s. After sometime its velocity becomes $8\widehat i + 20\widehat k$m/s due to the action of a conservative force. The work done on the particle during this interval of time is
(a) 0.32J
(b) 6.9J
(c) 9.6J
(d) 0.96J

Answer 1

Hint: If a conservative force acts on a particle then the total work done for a particle to displace from one point to another is independent of the path taken. Work done will simply be the difference between the kinetic energies of the two points.

Formula Used:
1. Kinetic energy\[K.E. = \dfrac{1}{2}m{v^2}\] ……(1)
Where,
m is the mass of the object
v is the velocity of the object
2. Work done: $W = \Delta K.E.$ ……(2)


Complete step by step answer:
Given:
1. Mass of the object m=0.01kg
2. Initial velocity of the particle ${v_i} = 4\widehat i + 16\widehat k$m/s
3. Final velocity of the particle ${v_f} = 8\widehat i + 20\widehat k$m/s

To find: The work done on the particle during this interval of time.

Step 1:
Find initial Kinetic energy using eq (1):
\[
  K.E.(i) = \dfrac{1}{2}m{v_i}^2 \\
   \Rightarrow K.E.(i) = \dfrac{1}{2}0.01 \times {(4\widehat i + 16\widehat k)^2} \\
 \]
Both the components get squared separately:
\[
   \Rightarrow K.E.(i) = \dfrac{1}{2}0.01 \times ({4^2} + {16^2}) \\
   \Rightarrow K.E.(i) = \dfrac{1}{2}0.01 \times 272 \\
   \Rightarrow K.E.(i) = 1.36J \\
 \]

Step 2:
Find final Kinetic energy using eq (1):
\[
   \Rightarrow K.E.(f) = \dfrac{1}{2}m{v_f}^2 \\
   \Rightarrow K.E.(f) = \dfrac{1}{2}0.01 \times {(8\widehat i + 20\widehat k)^2} \\
 \]
Both the components get squared separately:
\[
   \Rightarrow K.E.(f) = \dfrac{1}{2}0.01 \times ({8^2} + {20^2}) \\
   \Rightarrow K.E.(f) = \dfrac{1}{2}0.01 \times 464 \\
   \Rightarrow K.E.(f) = 2.32J \\
 \]

Step 3:
Use eq (2) to find the work done:
$
   \Rightarrow W = \Delta K.E. = K.E.(f) - K.E.(i) \\
   \Rightarrow W = 2.32 - 1.36 \\
   \Rightarrow W = 0.96J \\
 $

Final Answer
The work done on the particle during this interval of time is (d) 0.96J.


Note: Square of a vector is equal to taking its dot product with itself. So, we just square the components of the vector separately and add them:\[
  {(x\widehat i + y\widehat j + z\widehat k)^2} = (x\widehat i + y\widehat j + z\widehat k).(x\widehat i + y\widehat j + z\widehat k) \\
   \Rightarrow {(x\widehat i + y\widehat j + z\widehat k)^2} = ({x^2} + {y^2} + {z^2}) \\
 \]
 Square of a vector is equal to taking its dot product with itself. So, we just square the components of the vector separately and add them:\[
  {(x\widehat i + y\widehat j + z\widehat k)^2} = (x\widehat i + y\widehat j + z\widehat k).(x\widehat i + y\widehat j + z\widehat k) \\
   \Rightarrow {(x\widehat i + y\widehat j + z\widehat k)^2} = ({x^2} + {y^2} + {z^2}) \\
 \]
The dot product of velocity with itself is nothing but the square of the mod of the velocity vector.