Question

A body of mass 1 kg is moving with initial velocity of 1 m/sec is acted upon by a force for 1 sec. The increase in kinetic energy is:
A. 1 J
B. 2 J
C. 2.5J
D. 3 J

Answer 1

Answer:

1.5 J if force acting on particle is 1 N

Explanation:

Initial Kinetic energy

= 0.5mu²

= 0.5 × 1 kg × (1 m/s)²

= 0.5 J

Assuming the force of 1 N is acted upon the particle, the acceleration attained by particle is therefore

a = F/m

= 1 N / (1 kg)

= 1 m/s²

Velocity of particle after 1 s

v = u + at

= 1 m/s + (1 m/s² × 1 s)

= 2 m/s

Final Kinetic energy

= 0.5mv²

= 0.5 × 1 kg × (2 m/s)²

= 2 J

Increase in Kinetic energy = 2 J - 0.5 J = 1.5 J

Answer 2

Answer:

The increase in kinetic energy is 1.5J i.e.none of the options is correct.

Explanation:

The increase in kinetic energy is calculated as,

$E=\frac{1}{2}m(v^2-u^2)$     (1)

Where,

E=increase in kinetic energy of the body

m=mass of the body

v=final velocity

u=initial velocity

From the question we have,

Mass of the body=1 kg

The initial velocity(u)=1m/s

The time during which the force acts(t)=1 second

The force is calculated as,

$F=\frac{mu}{t}$         (2)

$F=ma$        (3)

a=acceleration of the body

By equating equations (2) and (3) we get;

$\frac{mu}{t}=ma$

$a=\frac{u}{t}$     (4)

By substituting the values in equation (4) we get;

$a=\frac{1}{1}$

$a=1m/s^2$      (5)

From the first equation of motion we have,

$v=u+at$      (6)

By substituting the values in equation (6) we have,

$v=1+1\times 1$

$v=2m/s$        (7)

By putting the values in equation (1) we get;

$E=\frac{1}{2}\times 1\times (2^2-1^2)$

$E=\frac{1}{2}\times 1\times (4-1)$

$E=\frac{1}{2}\times 1\times 3$

$E=1.5J$

Hence, the increase in kinetic energy is 1.5J i.e.none of the options is correct.

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