Learn work-kinetic energy theorem problems with examples and change in kinetic energy, practice equation to calculate work done by friction, gravity, constant force, spring force, force of gravity.
Thus, \(W = \Delta K E\)
This is known as work-Kinetic energy theorem.
As \((KE)_{final \;} >\;(KE)_{initial }\)
Then, work done,
\(W = \Delta K E\)
\(W = (KE)_{final}-(KE)_{initial }\)
\(W = Positive\)
As \((KE)_{final}\; <\; (KE) _{initial }\)
Then, work done, \(W = \Delta KE\)
\(W = (KE)_{final } - (KE) _{initial }\)
\(W = Negative \)
A Positive
B Negative
C Zero
D Increasing
Case-1: When a particle is projected in vertically upward direction
Change in kinetic energy
\(\Delta KE = (KE)_f - (KE)_i\)
\(\Delta KE = \dfrac {1}{2} × m × v_f^2- \dfrac{1}{2}× m ×v_i^2 \)
\(\Delta KE= \dfrac{1}{2}× m × (0)^2 - \dfrac{1}{2}× mv^2\)
\(\Delta KE = \dfrac{-1}{2} mv^2\)
\(W = \Delta KE\)
\(W = \dfrac{-1}{2}mv^2\)
Thus, work done is negative when particle is thrown vertically upward.
Case-2: When particle is dropped (free fall)
Change in kinetic energy
\(\Delta KE = (KE)_f- (KE)_i \)
\(\Delta KE = \dfrac{1}{2} m (v_f)^2-0\)
\(\Delta KE = \dfrac{1}{2} mv^2 \)
By work-energy theorem
W = \(\Delta\) KE
W = \(\dfrac{1}{2}mv^2\)
\((KE)_f-(KE)_i\;=\Delta KE\;[\because v_f= 0, v_i =0]\)
\(0\,-\;0\,=\,\Delta KE\)
\(\Delta KE = 0\)
W_{net} = W_{friction} + W_{gravity}
W_{net} = \(\Delta\) KE
W_{friction} + W_{gravity} = 0
W_{gravity }=_{ }–W_{friction}
\(mg\,sin\,\theta\times x=-(-\mu_kmg\,cos\,\theta)\)
\(x\,sin\,\theta=\mu _kcos\,\theta\)
By work energy theorem, W_{net} = \(\Delta KE\)
Work done by constant force + work done by spring force = \(\Delta KE\)
W_{c} + W_{s} = \(\Delta KE\)
W_{c} + W_{s }= 0
\(W_s=-W_c\)
(KE)_{i} = \(\dfrac{1}{2}\) mv^{2}
(KE)_{f} = \(\dfrac{1}{2}\) m(0)^{2 }= 0
W = \(\Delta\) KE
W = \(\left(\dfrac{1}{2}mv^2\right)_f-\left(\dfrac{1}{2}mv^2\right)_i\)
W = \(0-\dfrac{1}{2}mv_i^2\)
W = \(\dfrac{-1}{2}mv^2\)
\(F_s= -k\;x\)
Where negative sign shows the direction of spring force.
w = \(\int\limits^f_iF.dx\)
w = \(\int\limits^{x_f}_{x_i} - kx\;dx\)
w = – k \(\left[\dfrac{x^2}{2}\right]^{x_f}_{x_i}\)
w = \(\dfrac{-1}{2}k\;x_f^2\;+\dfrac{1}{2}k\;x_i^2\)
w = \(\dfrac{1}{2}k\;x_i^2\;-\;\dfrac{1}{2}k\;x_f^2\)
When \(x_f\;>\; x_i\) then, work done is negative.
When \(x_i\;>\; x_f\) then, work done is positive.
Case 1. Work done during extension
Here, the displacement (s) and force of spring (F_{s}) are in opposite direction. Thus, work done is negative.
Case 2. Work done during release after extension
Here, the displacement (s) and force of spring (F_{s}) are in same direction. Thus, work done is positive.
Case 3. Work done during compression
Here, the displacement (s) and force of spring (F_{s}) are in opposite direction. Thus, work done is negative.
Case 4. Work done during release after compression
Hence, the displacement (s) and force of spring (F_{s}) are in same direction. Thus, work done is positive.
A – 2.2 J
B 3.3 J
C – 4.4 J
D – 1.1 J