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NEET PHYSICSEasy

A particle moves from a point (−2i^+5j^)(-2\hat i+5\hat j) to (4j^+3k^)(4\hat j+3\hat k) when a force of (4i^+3j^)(4\hat i+3\hat j) N is applied. How much work has been done by the force?

A

8 J

B

11 J

C

5 J

D

2 J

Step-by-Step Solution

  1. Calculate Displacement Vector (d\mathbf{d}): The displacement is the difference between the final position vector (r2\mathbf{r}_2) and the initial position vector (r1\mathbf{r}_1). r1=−2i^+5j^\mathbf{r}_1 = -2\hat i + 5\hat j r2=4j^+3k^\mathbf{r}_2 = 4\hat j + 3\hat k d=r2−r1=(0−(−2))i^+(4−5)j^+(3−0)k^\mathbf{d} = \mathbf{r}_2 - \mathbf{r}_1 = (0 - (-2))\hat i + (4 - 5)\hat j + (3 - 0)\hat k d=2i^−1j^+3k^\mathbf{d} = 2\hat i - 1\hat j + 3\hat k
  2. Identify Force Vector (F\mathbf{F}): F=4i^+3j^\mathbf{F} = 4\hat i + 3\hat j
  3. Calculate Work Done (WW): Work is defined as the scalar (dot) product of force and displacement. W=F⋅dW = \mathbf{F} \cdot \mathbf{d} [Class 11 Physics, Ch 5, Sec 5.3, Eq 5.4] W=(4i^+3j^+0k^)⋅(2i^−1j^+3k^)W = (4\hat i + 3\hat j + 0\hat k) \cdot (2\hat i - 1\hat j + 3\hat k) Using the property i^⋅i^=1,i^⋅j^=0\hat i \cdot \hat i = 1, \hat i \cdot \hat j = 0, etc.: W=(4)(2)+(3)(−1)+(0)(3)W = (4)(2) + (3)(-1) + (0)(3) W=8−3+0=5 JW = 8 - 3 + 0 = 5\text{ J} [Class 11 Physics, Ch 5, Sec 5.1.1]
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