A 4 kg object has a velocity of 3.0 hat{i} | vedclass

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(A) Given: mass $m = 4 \ kg$, initial velocity $\vec{u} = 3 \hat{i} \ m/s$, final velocity $\vec{v} = (8 \hat{i} + 10 \hat{j}) \ m/s$, and time $t = 8

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$\frac{5 \sqrt{5}}{2} \ N$

Vedclass · Answer

(A) Given: mass $m = 4 \ kg$, initial velocity $\vec{u} = 3 \hat{i} \ m/s$, final velocity $\vec{v} = (8 \hat{i} + 10 \hat{j}) \ m/s$, and time $t = 8 \ s$.Using the first equation of motion, $\vec{v} = \vec{u} + \vec{a}t$, we find the acceleration $\vec{a}$:$\vec{a} = \frac{\vec{v} - \vec{u}}{t} = \frac{(8 \hat{i} + 10 \hat{j}) - 3 \hat{i}}{8} = \frac{5 \hat{i} + 10 \hat{j}}{8} \ m/s^2$.Now, using Newton's second law, $\vec{F} = m\vec{a}$:$\vec{F} = 4 	imes \left( \frac{5 \hat{i} + 10 \hat{j}}{8} ight) = \frac{1}{2} (5 \hat{i} + 10 \hat{j}) = (2.5 \hat{i} + 5 \hat{j}) \ N$.The magnitude of the force is $|\vec{F}| = \sqrt{(2.5)^2 + 5^2} = \sqrt{6.25 + 25} = \sqrt{31.25} = \sqrt{\frac{125}{4}} = \frac{5 \sqrt{5}}{2} \ N$.

$A$ $4 \ kg$ object has a velocity of $3.0 \hat{i} \ m/s$ at some instant. Eight seconds later, its velocity is $(8.0 \hat{i} + 10.0 \hat{j}) \ m/s$. Assuming that the object is subjected to a constant net force, the magnitude of the force is

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