Question

certain force acting on a 20 kg mass changes its velocity from 5 m/s to 2 m/s. calculate the workdone by
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Answer 1

Given: u = 5ms -¹

v = 2ms -¹

From work energy theorem,

$\large\bf { W = ∆K.E }$

$\large\bf { W = \frac { 1}{2} mv^{2} - \frac {1}{2} mv^{2}}$

$\large\bf { W = \frac {1}{2} \times 20 × 2² - \frac {1}{2} × 20 × 5²}$

$\large\bf { -210 J}$

Answer 2

CORRECT QUESTION:

Certain force acting on a 20 kg mass changes its velocity from 5 m/s to 2 m/s. calculate the workdone by the force.

ANSWER:

• The workdone by the force = - 210 J.

GIVEN:

• Mass of the body = 20 kg.

• Initial velocity = 5 m/s.

• Final velocity = 2 m/s.

TO FIND:

• The workdone by the force.

EXPLANATION:

$\boxed{ \bold{ \large{ \gray{Work\ done = Final \ K.E- Initial \ K.E }}}}$

$\boxed{ \bold{ \large{ \gray{K.E = \dfrac{1}{2}mv^2}}}}$

$\sf \mapsto Final \ K.E = \dfrac{1}{2}mv_f^2$

$\sf \mapsto Initial \ K.E = \dfrac{1}{2}mv_i^2$

$\sf \mapsto Work\ done = \dfrac{1}{2}mv_f^2 - \dfrac{1}{2}mv_i^2$

$\sf \mapsto Work\ done = \dfrac{1}{2}m(v_f^2 - v_i^2)$

$\sf \mapsto Work\ done = \dfrac{1}{2}(20)(2^2 - 5^2)$

$\sf \mapsto Work\ done = 10(4 - 25)$

$\sf \mapsto Work\ done = 10 (- 21)$

$\sf \mapsto Work\ done = - 210 \ J$

Hence the workdone by the force = - 210 J.