The potential energy of a particle of mass 5 | vedclass

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Question

(B) The force acting on the particle is $\vec{F} = -
abla V = -(\frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j}) = -(-7

Vedclass · Accepted Answer

the magnitude of acceleration of the particle is $5 \ m/s^2$

Vedclass · Answer

(B) The force acting on the particle is $\vec{F} = -
abla V = -(\frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j}) = -(-7 \hat{i} + 24 \hat{j}) = 7 \hat{i} - 24 \hat{j} \ N$.The acceleration is $\vec{a} = \frac{\vec{F}}{m} = \frac{7 \hat{i} - 24 \hat{j}}{5} = 1.4 \hat{i} - 4.8 \hat{j} \ m/s^2$.The magnitude of acceleration is $|\vec{a}| = \sqrt{1.4^2 + (-4.8)^2} = \sqrt{1.96 + 23.04} = \sqrt{25} = 5 \ m/s^2$. (Option $B$ is correct).At $t = 4 \ s$,the velocity is $\vec{v} = \vec{v}_0 + \vec{a}t = (1.44 \hat{i} + 4.2 \hat{j}) + (1.4 \hat{i} - 4.8 \hat{j}) 	imes 4 = (1.44 + 5.6) \hat{i} + (4.2 - 19.2) \hat{j} = 7.04 \hat{i} - 15 \hat{j} \ m/s$.The magnitude is $|\vec{v}| = \sqrt{7.04^2 + (-15)^2} \approx \sqrt{49.56 + 225} \approx 16.57 \ m/s$. (Option $A$ is incorrect).For option $C$,check the dot product $\vec{v}_0 \cdot \vec{a} = (1.44)(1.4) + (4.2)(-4.8) = 2.016 - 20.16 
eq 0$. (Option $C$ is incorrect).Since only $B$ is correct,the answer is $B$.

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