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`        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

Arun
23811 Points
```							Given,m=4 kg,S1=5m,S2=2mDistance traveled in nth second S=u+1/2a(2n-1)S1=5=u+1/2a(2(1)-1)=u+1/2a_______1S2=2=u+1/2a(2(3)-1)=u+5/2a———2From 1 and 2,We get a=-1.5m/s²,u=3.75m/sF=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 bySn=u + (a/2)(2n-1)-----------(1)Hence according to question, S1=5m and S3=2mFrom (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 = 4kgLet force=FHence by 2nd law of motion,F = maor F = 4(-3/2)or F = -6NHence 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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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions