Question

A 4.0-kg particle has an initial velocity of (3i – 4j) m/s. Some time later, its velocity is (7i + 3j) m/s. How much work was done by the resultant force during this time interval, assuming no energy is lost in the process?

Answer 1

Given:

A 4.0-kg particle has an initial velocity of (3i – 4j) m/s. Some time later, its velocity is (7i + 3j) m/s.

To Find:

How much work was done by the resultant force during this time interval, assuming no energy is lost in the process?

Solution:

$\mathbf{Initial\ velocity = \vec{u}=3\hat{i}-4\hat{j} }$

$\mathbf{Initial\ speed = u=\sqrt{3^{2}+4^{2}}=5 \ \ \frac{m}{s}}$

$\mathbf{Final\ velocity = \vec{v}=7\hat{i}+3\hat{j} }$

$\mathbf{Final\ speed = v=\sqrt{7^{2}+3^{2}}=\sqrt{58} \ \ \frac{m}{s}}$

Mass of object = M = 4.0 kg

We know that work done on object is equal to change in kinetic energy:

$\mathbf{Work\ done=\dfrac{1}{2}\times M\times (v^{2}-u^{2})}$

On putting respective value in above equation:

$\mathbf{Work\ done=\dfrac{1}{2}\times 4.0\times ((\sqrt{58})^{2}-5^{2})}$

On simplify:

Work done = 66 J

Means Work done by the resultant force during this time interval is 66 Joule.

Answer 2

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Means Work done by the resultant force during this time interval is 66 Joule.