A force of 4.0 n acts on a body of mass 2.0 kg for 4 s
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Applied force F=4N
The mass of the body to which the force is applied is m=2kg
By Newton's law acceleration produced a=Fm=42=2ms−2
The initial velocity of the body u=0
Tthe distance covered in first phase i.e. t=4 sec after the force starts acting will be
S1=u×t+12a×t2=0×4+12×2×42=16m
The forces ceases to act after 4 sec and the body moves with uniform velocity attained after 4sec.
The velocity attained v=u+a×t=0+2×4=8ms−1
So the distance covered in 2nd phase i.e. in 6sec will be
S2=v×6=8×6=48m
So total distance covered in 10sec will be
Stotal=S1+S2=16+48=64m
The mass of the body to which the force is applied is m=2kg
By Newton's law acceleration produced a=Fm=42=2ms−2
The initial velocity of the body u=0
Tthe distance covered in first phase i.e. t=4 sec after the force starts acting will be
S1=u×t+12a×t2=0×4+12×2×42=16m
The forces ceases to act after 4 sec and the body moves with uniform velocity attained after 4sec.
The velocity attained v=u+a×t=0+2×4=8ms−1
So the distance covered in 2nd phase i.e. in 6sec will be
S2=v×6=8×6=48m
So total distance covered in 10sec will be
Stotal=S1+S2=16+48=64m
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