Question

a body of mass 5 kg is pulled by a force F=(-4i+3j) N from rest on a smooth horizontal table in x z plane the time at which it will attain a speed of 4 ms^-1... ans is 5 s how??​

Answer 1

a body of mass 5kg pulled by a force F = -4i + 3j

according to Newton's 2nd law,

F = ma

so, a = F/m = (-4i + 3j)/5 = -0.8i + 0.6j

hence, acceleration , a = -0.8i + 0.6j

but according to question, body is pulled from rest on a smooth horizontal table in xz plane.

so, acceleration along xz plane , a' = -0.8i

magnitude of a' = 0.8 m/s²

now, use formula, v = u + at

4 = 0 + 0.8t

t = 5 sec

Answer 2

Answer: The body will attain a speed at 5 sec.

Explanation:

Given that,

Mass of a body m = 5 kg

Force $F=(-4i+3j) N</p><p>Speed v = 4 m/s</p><p><strong>Using newton's second law</strong></p><p>[tex]F = ma$

$a = \dfrac{F}{m}$

$a = \dfrac{-4i+3j}{5}$

$a = (-0.8i+0.6j) m/s^2$

According to question,

A body pulled from rest on a smooth horizontal table in x z plane.

The magnitude of the acceleration will be -0.8 in horizontal x-z plane

Using equation of motion

$v = u-at$

$4 = 0+0.8t$

$t = 5 sec$

Hence, The body will attain a speed at 5 sec.