Question

Certain force acting on a 30 kg mass changes its velocity from 12 ms^-1 to 15ms^-1.Calculate the work done by the force.

Answer 1

To Find :-

• The work done by the force.

Solution :-

• Initial velocity (u) = 12 ms-¹

• Final velocity (v) = 15 ms-¹

• Mass (m) = 30 kg

According to the Work - Energy theorem,

● Work done = Change in kinetic energy

$\red\bigstar$ W = ΔK.E

[ Put the values ]

$\longrightarrow \sf \: W = \frac{1}{2} m(v {}^{2} - u {}^{2} )$

$\longrightarrow \sf \: W = \frac{1}{2} \times 30(15 {}^{2} - 12 {}^{2} )$

$\longrightarrow \sf \: W = 15(225- 144)$

$\longrightarrow \sf \: W = 15(81)$

$\longrightarrow \sf \: W = 1215 \: joule \: \: \green \bigstar$

Therefore,

The work done by the force is 1215 J.

Answer 2

Explanation:

Given:-

initial velocity =u=${12ms}^{-1}$

final velocity =v=${15ms}^{-1}$

mass=m=30kg

To find:-

work done by force=W

Solution :-

as we know

${:}\longrightarrow$${\boxed {work\:done=change \:in\:kinetic\:Energy}}$

${:}\longrightarrow$${\boxed{W={\triangle}K.E}}$

${:}\longrightarrow$${\frac{1}{2}}m ({v}^{2}-{u}^{2})$

${:}\longrightarrow$${\frac {1}{{\cancel{2}}}}×{\cancel{30}}×({15}^{2}-{12}^{2})$

${:}\longrightarrow$$15(225-144)$

${:}\longrightarrow$$15×81$

${:}\longrightarrow$$1215Joule$

${:}\longrightarrow$${\underline{\boxed{\bf {W=1215J}}}}$