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Grade: 12
        
a body of mass 4 kg is accelerated upon by a constant force travels a distance of 5m in first second and a distance of 2m in the third second . the force acting on the body?
a] 6N    b]8N   c]2N    d]4N
4 months ago

Answers : (2)

Arun
23811 Points
							

Given,m=4 kg,S1=5m,S2=2m

Distance traveled in nth second S=u+1/2a(2n-1)

S1=5=u+1/2a(2(1)-1)=u+1/2a_______1

S2=2=u+1/2a(2(3)-1)=u+5/2a———2

From 1 and 2,

We get a=-1.5m/s²,u=3.75m/s

F=ma=4*(-1.5)=-6kN

4 months ago
Vikas TU
10573 Points
							
Dear student 
As the force acting on the body is constant, its acceleration will also be constant.
Let initial velocity='u'
Let acceleration='a'
The distance travelled by a body undergoing consant acceleration in the nth second is given by
Sn=u + (a/2)(2n-1)-----------(1)
Hence according to question, S1=5m and S3=2m
From (1),
S1=u + (a/2)[2(1)-1]
or 5=u + a/2----------(2)
Also, S3=u + (a/2)[2(3)-1]
or 2=u + (5/2)a-----------(3)
Solving (2) and (3) for 'u' and 'a' we get,
u=33/6 and a=(-3/2)
Hence acceleration=a=-3/2 ms-2.
Mass of body = m = 4kg
Let force=F
Hence by 2nd law of motion,
F = ma
or F = 4(-3/2)
or F = -6N
Hence the magnitude of force is 6N and is acting opposite to the direction of motion of the body.
4 months ago
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