Question

certain force acting on a 20 kg mass changes it's velocity fro 5 m/S to 2 m/S. calculate the work done by the force

Answer 1

Answer:

-160 J

Explanation:

Work - Energy Theorem states that the work done on the body by the force in the direction of motion is the change in the kinetic energy.

The relation -

$Work = \frac{1}{2}m(v^{2} - u^{2} )$

Mass given - 20 kg

Initial velocity - 5m/s

Final velocity - 2m/s

$Work = \frac{1}{2}X 20 X(2^{2} - 5^{2})\\= 10 X (4-25) \\= 10 X (-16)\\= -160 J$

The work done on the body is -160 J

Answer 2

We have,

Mass of object = 20 kg

Initial velocity = 5 m/s

Final velocity = 2 m/s

We know, the work - energy theorem states that,

$\boxed{W = ∆KE = \frac{1}{2}(v + u)(v - u)}$

⇒ W = ½(20 kg)(2 m/s + 5 m/s)(2 m/s - 5 m/s)

⇒ W = (10 kg)(7 m/s)(- 3 m/s)

⇒ W = - 210 Nm

⇒ W = - 210 J (answer).

More:-

• Work done is a scalar quantity. But that doesn't mean that it is always Positive.
• It's SI or MKS unit is "Joule" (J). And it's CGS unit is "erg".
• It's dimension formula is $M^1 L^2 T^{-2}$.