A force of $2\hat{i}+3\hat{j}+4\hat{k}N$ acts on a body for 4 second, produces a displacement of $(3\hat{i}+4\hat{j}+5\hat{k})m.$ The power used is-

(1) 9.5 W

(2) 7.5 W

(3) 6.5 W

(4) 4.5 W

If \(\vec{F} = (60\hat{i} + 15\hat{j}-3\hat{k})\) N and \(\vec{v} = (2\hat{i} - 4\hat{j}+5\hat{k})\) m/s, then instantaneous power is:
1. \(195\) watt
2. \(45\) watt
3. \(75\) watt
4. \(100\) watt

A constant force $\overrightarrow{\mathrm{F}}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\left(2\hat{\mathrm{i}}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>3\hat{\mathrm{j}}<mspace\; width="1em"></mspace>-<mspace\; width="1em"></mspace>\hat{\mathrm{k}}\right)$ acts on a particle. At a certain instant when velocity of the particle is $\overrightarrow{\mathrm{v}}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\left(\mathrm{a}\hat{\mathrm{i}}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\mathrm{b}\hat{\mathrm{j}}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\hat{\mathrm{k}}\right)$${\mathrm{ms}}^{-1}$. The power of the force is zero, then

(1)  2a + 3b = 1

(2)  a + b = 1

(3)  2a = 36

(4)  a + b + 1 = 0

A force $\overrightarrow{F}=6\hat{i}+2\hat{j}-3\hat{k}$ acts on a particle and produces a displacement of $\overrightarrow{s}=2\hat{i}-3\hat{j}+x\hat{k}.$ If the work done is zero, the value of x is 

(1) – 2

(2) 1/2

(3) 6

(4) 2

A particle moves from position ${\overrightarrow{r}}_1=3\hat{i}+2\hat{j}-6\hat{k}$ to position ${\overrightarrow{r}}_2=14\hat{i}+13\hat{j}+9\hat{k}$ under the action of force $4\hat{i}+\hat{j}+3\hat{k}N.$ The work done will be 

(1) 100 J

(2) 50 J

(3) 200 J

(4) 75 J

A motor pulls a block by giving a force of 50 N at a speed of 36 km/h. The power supplied by the motor to the block is:

1.  500 watt

2.  1800 watt

3.  250 watt

4.  200 watt