Question

Certain force acting on a 20 kg mass changes it's velocity from 5 ms-1 to 2 ms -1. Calculate the work done by the force.

Answer 1

Mass, m = 20 kg

Initial velocity, u = 5 m/s

Final velocity, v = 2 m/s

Time, t = 1s

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

From third equation of motion,
 

v² - u² = 2as
⇒ (2ms-1)² - (5ms-1)² = 2 x (-3ms-²) x s



⇒          -21 m²s-² = -6 ms-² x s

⇒ s = 7/2 m

Work done = F x s

Force = m x a

Therefore,
 
W = m x a x s
    = 20 kg x (-3ms-²) x 7/2 m
   = -210 Joules
 

Answer 2

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

therefore, we know ,
workdone= f × s
and , F = m× a

so, work = m× a× s
= 20 kg × ( -3ms-²) × 7/2m
= - 210 joules.