A particle moves with the velocity `vecv=(5hati+2hatj-hatk)ms^(-1)` under the influence of a constant force, `vecF=(2hati+5hatj-10hatk)N.` The instantaneous power applied is

Options

A. 5 W
B. 10 W
C. 20 W
D. 30 W

Answer

To find the instantaneous power applied to the particle, we can use the formula for power in terms of force and velocity. The instantaneous power \( P \) is given by the dot product of the force vector \( \vec{F} \) and the velocity vector \( \vec{v} \):

\[ P = \vec{F} \cdot \vec{v} \]

### Step 1: Write down the force and velocity vectors

The force vector is given as:
\[ \vec{F} = 2\hat{i} + 5\hat{j} - 10\hat{k} \, \text{N} \]

The velocity vector is given as:
\[ \vec{v} = 5\hat{i} + 2\hat{j} - 1\hat{k} \, \text{m/s} \]

### Step 2: Calculate the dot product \( \vec{F} \cdot \vec{v} \)

The dot product of two vectors \( \vec{A} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \) and \( \vec{B} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \) is calculated as:
\[ \vec{A} \cdot \vec{B} = a_1b_1 + a_2b_2 + a_3b_3 \]

Applying this to our vectors \( \vec{F} \) and \( \vec{v} \):
\[ \vec{F} \cdot \vec{v} = (2)(5) + (5)(2) + (-10)(-1) \]

### Step 3: Perform the calculations

Calculating each term:
- First term: \( 2 \times 5 = 10 \)
- Second term: \( 5 \times 2 = 10 \)
- Third term: \( -10 \times -1 = 10 \)

Now, sum these results:
\[ 10 + 10 + 10 = 30 \]

### Step 4: Conclusion

The instantaneous power applied to the particle is:
\[ P = 30 \, \text{W} \]

Thus, the answer is \( 30 \, \text{W} \).

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