A 0.2 kg object at rest is subjected to a force (0.3 hati-0.4 hat j) N. What is the velocity after 6 s?

Question

A 0.2 kg object at rest is subjected to a force (0.3 hati-0.4 hat j) N. What is the velocity after 6 s? (A) (9 hati-12 hat j) (B) (8 hati-16 hat j) (C) (12 hati-9 hat j) (D) (16 hati-8 hat j)

Tardigrade Admin · Accepted Answer

Here, m = 0.2 kg, u = 0 vecF=(0.3 hati-0.4 hatj) vecv = ?, t = 6 s veca=( vecF/m)=((0.3 hati-0.4 hatj)/0.2)=((3/2) hati-2 hatj ) From vecv= vecu+ veca t vecv=0+((3/2) hati-2 hatj ÷ )×6=9 hati-12 hatj

Q. A 0.2 kg object at rest is subjected to a force $(0.3 \hat{i} - 0.4 \hat{j})$ N. What is the velocity after 6 s?

$(9 \hat{i} - 12 \hat{j})$

$(8 \hat{i} - 16 \hat{j})$

$(12 \hat{i} - 9 \hat{j})$

$(16 \hat{i} - 8 \hat{j})$

Here, $m = 0.2 k g, u = 0$
$F = (0.3 \hat{i} - 0.4 \hat{j})$
$v$ = ?, $t = 6 s$
$a = \frac{F}{m} = \frac{( 0.3 i ^ - 0.4 j ^ )}{0.2} = (\frac{3}{2} \hat{i} - 2 \hat{j})$
From $v = u + a t$
$v = 0 + (\frac{3}{2} \hat{i} - 2 \hat{j} \div) \times 6 = 9 \hat{i} - 12 \hat{j}$

Q. A 0.2 kg object at rest is subjected to a force (0.3i^−0.4j^​) N. What is the velocity after 6 s?

(9i^−12j^​)

(8i^−16j^​)

(12i^−9j^​)

(16i^−8j^​)