Question

A force of 1 N acts on a 1 kg mass for 1 second which is initially at rest. In another case 1 N force acts on 1 kg mass which is initially at rest and moves it through 1 m. The ratio of kinetic energy in the two cases is

(1) 1:1 (3) 1:4

(2) 1:2

(4) 1:√2​

Answer 1

Answer:

F=1N,t=1S

△P=F×t=1N×1S=1kgm/s

P

f

−P

i

=1kgm/s

P

f

−0=1kgm/s

P

f

=1kgm/s

F=1N,M=1kgm,S=1m

a=

M

F

,

1Kg

1N

=1m/s

2

V

f

2

−V

i

2

=2as

V

f

2

−0=2(1m/s

2

)(1m)

V

f

=

2ms

multiply both sides

mv

f

=

2m/s

×1kg

P

′

=

2

kgm/s

p

′

p

=

2

1

=

2

1

=1:

2

Answer 2

The ratio of kinetic energy in the two cases is 1 : √2

Explanation:

GIVEN

• F = 1N
• t = 1S

We know that

• △P = F×t = 1N×1S = 1kgm/s

$\rm \: P_f \: − \: P_i =1kgm/s \\ \rm P_f \: − \: 0=1kgm/s \\ \rm \: P_f =1kgm/s$

Now

• F = 1N
• M = 1kgm
• S = 1m

$\bf \: a = \cfrac {F}{ M} = \cfrac{1N}{1Kg}=1m/s²$

$\displaystyle \cal V_f ^2 −V_i^2 =2as \\ \\ \displaystyle \cal V_f^2 −0=2(1m/s² )(1m) \\ \\ \displaystyle \cal V_f = 2ms$

Multiply Both Sides

$\sf \: mv_f = \sqrt{2m/s} ×1kg \\ \\ \sf \: P ′ = \sqrt{2} kgm/s \\ \\ \bf \: \frac{P}{P′} = \frac{1}{\sqrt{2}} =1: √2$