A force 2 hat i + hat j - hat k newton acts on a body a which is initially at rest. If the velocity of the body at the end of 20 seconds is 4 hati+2 hatj+2 hatk ms-1 the mass of the body.

Question

A force 2 hat i + hat j - hat k newton acts on a body a which is initially at rest. If the velocity of the body at the end of 20 seconds is 4 hati+2 hatj+2 hatk ms-1 the mass of the body. (A) 20kg (B) 15kg (C) 10kg (D) 5kg

Tardigrade Admin · Accepted Answer

Considering that at the end of 20 sec, velocity is given by

4 \hat{i} + 2 \hat{j} + 2 \hat{k}

. therefore acceleration is given by,

a = v / t

.

a = \frac{4 i ^ + 2 j ^ + 2 k ^}{20}

Now using

m = F / a

and

F = 2 \hat{i} + \hat{j} + \hat{k}

Hence,

m = 10 k g

Q. A force 2i^+j^​−k^ newton acts on a body a which is initially at rest. If the velocity of the body at the end of 20seconds is 4i^+2j^​+2k^ms−1 the mass of the body.

20kg

15kg

10kg

5kg

Considering that at the end of 20 sec, velocity is given by 4i^+2j^​+2k^. therefore acceleration is given by, a=v/t. a=204i^+2j^​+2k^​ Now using m=F/a and F=2i^+j^​+k^ Hence, m=10kg

Q. A force $2 \hat{i} + \hat{j} - \hat{k}$ newton acts on a body a which is initially at rest. If the velocity of the body at the end of $20 seco n d s$ is $4 \hat{i} + 2 \hat{j} + 2 \hat{k} m s^{- 1}$ the mass of the body.

Considering that at the end of 20 sec, velocity is given by $4 \hat{i} + 2 \hat{j} + 2 \hat{k}$ . therefore acceleration is given by, $a = v / t$ . $a = \frac{4 i ^ + 2 j ^ + 2 k ^}{20}$
Now using $m = F / a$ and $F = 2 \hat{i} + \hat{j} + \hat{k}$
Hence, $m = 10 k g$