Question

A 0.2kg object at rest is subjected to a force 0.3-0.4j Newton. What will be it's velocity after 6sec

Answer 1

Answer:

Here, m = 0.2 kg, u=0

F=(0.3i - 0.4j)

v= ? , t = 6s

a = F/m

= (0.3i - 0.4j)/ 0.2

= (3/2i - 2j)

From v = u +at,

v = 0+(3/2i - 2j)*6

= 9i - 12j

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Answer 2

Solution:

Here, m=0.2kg,u=0

F⃗ =(0.3i^−0.4j^)

v⃗ = ?, t=6s

a⃗ =F⃗ m=(0.3i^−0.4j^)0.2=(32i^−2j^)

From v⃗ =u⃗ +a⃗ t

v⃗ =0+(32i^−2j^÷)×6=9i^−12j^