Question

A body of mass 6kg is acted upon a constant force F=(2i-3j)N. If it's initial velocity at t=0 is v=(-3i+4j)m/s, then thw time at which its will be along the X-axis is

Answer 1

Answer:

t = 8 sec

Explanation:

F = (2i - 3j) N

U = -3i + 4j

V = xi ( as it's along x-axis, y=0)

0 = F/m = (2i - 3j) / 6

V- U = at

By rearranging,

v= u + at

xi = -3i + 4j + 2i - 3j/6

As j component on L.H.S is zero. Then,

0 = 4 - 3t/6

t = 8 sec

Answer 2

Answer:

The time will be 8 sec.

Explanation:

Given that,

Mass m = 6 kg

Force $F-= (2i-3j)\ N$

Initial velocity $u = (-3i+4j)\ m/s$

Final velocity will be xi because it velocity is in x-direction so the y component  will be zero.

Final velocity $v = (xi)$

Using newton's second law

$F = ma$

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

$a = \dfrac{(2i-3j)}{6}\ m/s^2$

Using equation of motion

$v=u+at$

$xi=(-3i+4j)+\dfrac{(2i-3j)}{6}t$

y-component is zero in L.H.S

The time will be

$0=4+\dfrac{-3t}{6}$

$t = 8\ sec$

Hence, The time will be 8 sec.