Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • A body of mass 10 kg is acted upon by two perpendicular forces, 6N and 8N. The resultant acceleration of the body is

A body of mass 10 kg is acted upon by two perpendicular forces, 6N and 8N. The resultant acceleration of the body is
(a) 1 m s^{-2}at an angle of \tan^{-1}\left ( \frac{4}{3} \right ) w.r.t. 6N force.
(b) 0.2 ms^{-2} at an angle of \tan^{-1}\left ( \frac{4}{3} \right ) w.r.t. 6N force.
(c) 1 ms^{-2} at an angle of \tan^{-1}\left ( \frac{3}{4} \right )w.r.t. 8N force.
(d) 0.2 ms^{-2} at an angle of \tan^{-1}\left ( \frac{3}{4} \right ) w.r.t. 8N force.

Answers (1)

The correct answer is:

(a) 1 m s^{-2}at an angle of \tan^{-1}\left ( \frac{4}{3} \right ) w.r.t. 6N force.

(c) 1 ms^{-2} at an angle of \tan^{-1}\left ( \frac{3}{4} \right )w.r.t. 8N force.

Explanation:

Here we know that,

 m = 10kg, F2 = 8N & F1 = 6N

R =\sqrt{ 8^{2}+6^{2}}

R =\sqrt{ 64+36}

R =\sqrt{ 100}

Thus, R = 10 N

F = ma \rightarrow a = \frac{F}{m} = \frac{R}{m} = \frac{10}{10} = 1ms^{-2}      ………. (i)

tan \theta_{1} =\frac{8}{6} = \frac{4}{3}=\theta_{1} = \tan^{-1}\left (\frac{4}{3} \right )   ………… (ii)

tan \theta_{2} =\frac{6}{8} = \frac{3}{4}=\theta_{2} = \tan^{-1}\left (\frac{3}{4} \right )       ……….. (iii)

Hence, opt (a) is verified from (i) and (ii), whereas opt (c) is verified from (i)  and (iii).

Since, acceleration is not equal to 0.2 ms-2, opt (b) & (d) are rejected.

Posted by

infoexpert24

View full answer

Similar Questions

Latest Question