Question

Certain force acting on a 20kg mass changes it's velocity from 5m)sec to 2m/sec. Calculate the work done by the force?​

Answer 1

Answer:

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Explanation:

here , given mass = 20 kg

initial velocity (u) = 5 m/sec

final velocity (v) = 2 m/sec

time (t)= 1sec

so , from first equation of motion

v= u+ at

or, a= ( v- u) / t

a= ( 2 ms-¹ - 5 ms-¹) / 15

a= - 3 ms-²

therefore, from third equation of motion

v²- u²= 2as

= ( 2ms -1 )² - ( 5ms-¹) ²

= 2× (-3ms-²) ×5

= -21 m²s² = -6ms-²× 5

= s = 7/2 metre =3.5 m

therefore, we know ,

workdone= f × s

and , F = m× a

so, work = m× a× s

= 20 kg × ( -3ms-²) × 7/2m

= - 210 joules.

Answer 2

The work done by the force is -210 J.

• Given data:

initial velocity($v_{0}$) = 5 m/s.

final velocity($v_{f}$) = 2 m/s.

Mass(M)= 20 Kg.

• we know that Work(w) is $W =F*d$.

• Force(F) = M * a.

• by equaltion of motion:

                             

                      $v_{0}^{2} -v_{f} ^{2} = 2 * a* d\\ 2^{2}-5^{2} = 2* W /20\\W =-21 *10\\W =-210 J$