Question

Three forces f1 vector =( 2i^+4j^ ) N; f2vector = (2j^- k^) N and f3 vector = (k^- 4j^ -2i^) N are applied on an object of mass 1 kg at rest at origin. the position of the object at t = 2s will be

Answer 1

Answer:

At t = 2 s the object is at a distance of 8 m from initial position.

Explanation:

$f_{1} ,  f_{2}, f_{3}$ all forces are applied on the object.

Thus, resultant force F = $f_{1} + f_{2} + f_{3}$

$f_{1} = 2\hat{i} + 4\hat{j}$

$f_{2} = 2\hat{j} - \hat{k}$

$f_{3} = -2\hat{i} - 4\hat{j}+ \hat{k}$

F = $2\hat{i} + 4\hat{j} + 2\hat{j} - \hat{k}-2\hat{i} - 4\hat{j}+ \hat{k}$

F = $2\hat{j}$

Thus, F = 2 N

According to Newton's second law of motion,

F = ma

m = 1 kg

F = 2 N

Thus, a = $\frac{F}{m}$

a = 2 $\frac{m}{s^{2} }$

Now, Velocity is given by,

v = at

v = 2 × 2

v = 4$\frac{m}{s}$

Distance is given by,

s = v t

s = 4 × 2

s = 8 m

Hence, at t = 2 s the object is at a distance of 8 m from initial position.

Answer 2

Answer:

(-4m,8m)

Explanation:

Refer the image given