Question

A body of mass 2 kg starts with an initial velocity
5 m/s. If the body is acted upon by a time
dependent force (F) as shown in figure, then work
done on the body in 20 s is​

Answer 1

we know, The area under the curve of a graph of force vs. time is the impulse.

so, impulse = 20N(5s) + 1/2 × (10s - 5s) × 20N + 1/2 × (20s - 10s) × (-20N)

= 100Ns + 50Ns - 100Ns

= 50Ns

as we know, impulse = change in linear momentum

⇒50 Ns = m(v - u)

⇒50 Ns = 2kg(v - 5m/s)

⇒v = 30 m/s

now, workdone on the body in 20s = final kinetic energy - initial kinetic energy

= 1/2 × 2kg × (30m/s)² - 1/2 × 2kg × (5m/s)²

= 900 J - 25 J

= 875 J

hence, option (2) is correct choice.

Answer 2

$\huge\bold\purple{Answer:-}$

impulse = 20N(5s) + 1/2 × (10s - 5s) × 20N + 1/2 × (20s - 10s) × (-20N)

= 100Ns + 50Ns - 100Ns

= 50Ns

as we know, impulse = change in linear momentum

⇒50 Ns = m(v - u)

⇒50 Ns = 2kg(v - 5m/s)

⇒v = 30 m/s

now, workdone on the body in 20s = final kinetic energy - initial kinetic energy

= 1/2 × 2kg × (30m/s)² - 1/2 × 2kg × (5m/s)²

= 900 J - 25 J

= 875 J

hence, option (2) is correct choice.