Question

a block of mass 3kg has a velocity of um/s. when a force of 18N acts on the block. it reduces the velocity from um/s to u/2 after the block has covered a distance of 9m. find u

Answer 1

hello there!!!!

ATQ,
mass = 3 kg, force= 18N
initial velocity= u m/ s
final velocity ,v = u/2 m/ s
As we know that F= ma
acc. to Newton's first law of motion...
F = ma
18 kg m/ s²=( 3kg )a. , a= 6 m/s²

As we know that v²- u² = 2as
S is given ie 9 m
V²- U²= 2AS
( U/2)² - U² = 2× 6 × 9
-3U²/4 = 108
-U² = 144
- U = +12 , -12
-U = -12
:......................................
: U= 12M/ S :::
:.....................................:
answer is
u = 12m/ s


HOPE IT HELPS YOU

Answer 2

F = 18N
mass= 3kg
initial velocity = u m/s
final velocity = u/2 m/s
distance covered(S) = 9 m
F = 18N
ma = 18
3(a) = 18
a = 6 m/s^2
Now, from Newton equation
v^2 = u^2 + 2aS
u^2/4 = u^2 + 2(6)(9)
3u^2/4 = 108
u^2 = 108*4/3
u^2 = 144
u =12 m/s