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Question: A 0.2 kg object at rest is subjected to a force (\(0.3 \hat{i} -0.4 \hat{j}\)) N. What is the veloci...

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

Explanation

Solution

To solve this problem, use the formula for force which gives relation between force and acceleration. Substitute the values and obtain the value for acceleration. Then, use Newton's law of motion giving the relation between initial velocity and final velocity. Substitute the values and calculate the value for velocity of the object after 6s.

Complete step-by-step answer:
Given: Force (F)= (0.3i^−0.4j^0.3 \hat{i} -0.4 \hat{j}) N
Time (t) = 6s
Mass (m)= 0.2 kg
Initial velocity (u)= 0
Let the velocity of the object after 6s be v.
We know, force is given by,
F=maF= ma …(1)
Where, m is the mass of the object
a is the acceleration
Rearranging equation. (1) we get,
a=Fma= \dfrac {F}{m}
Substituting the values in above equation we get,
a=(0.3i^−0.4j^)0.2a= \dfrac {(0.3 \hat{i} -0.4 \hat{j}) }{0.2}
⇒a=(32i^−2j^)\Rightarrow a = (\dfrac {3}{2} \hat{i} -2 \hat{j})
From, Newton’s law of motion we know,
v=u+atv= u + at
Substituting the values in above equation we get,
v=0+(32i^−2j^)v = 0 + (\dfrac {3}{2} \hat{i} -2 \hat{j})
⇒v=(32i^−2j^)×6\Rightarrow v = (\dfrac {3}{2} \hat{i} -2 \hat{j}) \times 6
⇒v=(182i^−12j^)\Rightarrow v = (\dfrac {18}{2} \hat{i} - 12 \hat{j})
⇒v=(9i^−12j^)\Rightarrow v = (9 \hat{i} - 12 \hat{j})
Hence, the velocity of an object after 6s is (9i^−12j^)(9 \hat{i} - 12 \hat{j}).
So, the correct answer is option A i.e. (9i^−12j^)(9 \hat{i} -12 \hat{j}).

So, the correct answer is “Option A”.

Note: The above mentioned problem can be solved using an alternate method. The alternate method is given below:
We know, force acting on an object is given by,
F=dpdtF = \dfrac {dp}{dt} …(1)
Where, dp is the change in the particle’s momentum
Dt is the change in the time
Rearranging equation. (1) we get,
dp=F.dtdp= F. dt
Integrating the above equation we get,
p=∫0tF.dtp = \int_{0}^{t} F .dt
⇒p=(0.3i^−0.4j^)×6\Rightarrow p = (0.3 \hat{i} -0.4 \hat{j}) \times 6
⇒p=(1.8i^−2.4j^)\Rightarrow p = (1.8 \hat{i} -2.4 \hat{j})
But, p=mvp= mv,
⇒mv=(1.8i^−2.4j^)\Rightarrow mv = (1.8 \hat{i} -2.4 \hat{j})
⇒v=(1.8i^−2.4j^)m\Rightarrow v = \dfrac {(1.8 \hat{i} -2.4 \hat{j})}{m}
⇒v=(1.8i^−2.4j^)0.2\Rightarrow v = \dfrac {(1.8 \hat{i} -2.4 \hat{j})}{0.2}
⇒v=(9i^−12j^)\Rightarrow v = (9 \hat{i} - 12 \hat{j})
Hence, the velocity of an object after 6s is (9i^−12j^)(9 \hat{i} - 12 \hat{j}).