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Grade: 9
        
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.
one year ago

Answers : (6)

Khimraj
3008 Points
							
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.
one year ago
Sangamesh Davey
28 Points
							
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
one year ago
Devansh sharma
14 Points
							
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
one year ago
Devansh sharma
14 Points
							sorry
						
one year ago
Ad
15 Points
							
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.
 
one year ago
Waishnavi Mahajan
17 Points
							
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
9 months ago
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