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# A force of 6N acts on a body of mass 1.5 Kg for 2 seconds. Assuming the body to be initially at rest, find the distance covered in 5sec after the force starts acting.

Khimraj
3007 Points
2 years ago
acceleration of body for 2 sec
a = 6/1.5 = 4 m/s2
velocity after 2 sec
v = a*t = 4*2 = 8 m/s
Distance coververd in 5 sec
d = ( ½*4*22) + (8*3) = 8+24 = 32 m
Hope it clears.
Sangamesh Davey
28 Points
2 years ago
F=ma
6=1.5 *a
a=4m/s2
distance covered when the body is at  rest = s= 1/2at2     (that is because u =0m/s)
Therefore ,
s=1/2 (4)(4)
=8m
Devansh sharma
14 Points
2 years ago
FORCE=mass*acc.---------1                                   (F=6N,,M=1.5kg,,t=5 sec)
s=ut+1/2at^2                                               ( we are given that body is initially at rest so,ut=0)
so,s=1/2 at^2----------------2
by,1 equation                             6=1.5*a
a=6/1.5=4m/s^2
by using equation 2
1/2*4*5^2
2*25
=50N
Devansh sharma
14 Points
2 years ago
sorry
15 Points
2 years ago
CASE-1
Given.
F=6N
m=1.5kg
t=2 seconds
*F=ma
F/m=a
6/1.5=4mps,sq.
Case-2
T=5 sec
U=0mps
S=ut+1/2a(t,sq.)
S=0×5+1/2×4×25
S=50m.

Waishnavi Mahajan
17 Points
2 years ago
F=6 N
M=1.5 kg
t=2sec
,,F=mu/t
,,u=ft/m=12/1.5
,,S=Ur+1/2at^2
S=12/1.5+ 1/2*6/1.5*(5)^2
S=40+50
S=90 m
Tanu
13 Points
11 months ago
We know ....F =ma
6=1.5 × a
6/1.5 =a
a= 4 m/sec square
We also know that .....v=u+at
v = 0+4×2
v =8 m / sec
We are aware that
S =ut+ 1/2 at square
S = 0×5 +1/2 ×4×2 ×2
S=8m
= 8m + the distance of the 3 sec
= 8+ 3×8
=8+24
=32m