Question

A 5.0 kg particle has 160 j of work done on it if it's initial speed is zero what is its final speed in m/s

Answer 1

$\Huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

From the Question,

• Work Done,W = 160 J

• Initial Velocity,u = 0 m/s

• Mass,m = 5 Kg

To find

Final Velocity of the particle

According to Work Energy Theorem,

$\sf{W = \frac{1}{2}m{v}^{2} - \frac{1}{2}m{u}^{2}} \\ \\ \rightarrow \large{\boxed{\boxed{ \sf{W = \frac{1}{2}m{v}^{2} }}}}$

Putting the values,we get:

$\sf{180 = \frac{1}{2}(5)v {}^{2} } \\ \ \\ \implies \: \sf{v {}^{2} = \frac{360}{5} } \\ \\ \implies \: \sf{v = \sqrt{180} } \\ \\ \implies \ \: \large{ \underline{ \boxed{ \sf{v = 8.5ms {}^{ - 1} }}}}$

Thus,the change in velocity is 8.5 m/s

Answer 2

Answer:

Explanation:

$\bf\underline{Given:-}$

Work Done of particle, w = 160 J

Initial Velocity of particle, u = 0 m/s

Mass of the particle, m = 5 Kg

$\bf\underline{To \: Find:-}$

Final speed

$\bf\underline{Formula \: to \: be \: used:-}$

${\boxed{\bf\underline {W=\frac{1}{2}mv^2}}}$

$\bf\underline{Solution:-}$

Putting all the values, we get

$\sf\implies W=\frac{1}{2}mv^2$

$\sf\implies 180=\frac{1}{2}\times5 \times v^2$

$\sf\implies v^2=\frac{360}{5}$

$\sf\implies v^2=\sqrt{180}$

$\bf\implies v=8.5 \: m/s$

Hence, the final velocity of particle is 8.5 m/s.