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A body of mas...

Question

A body of mass 8 kg is hanging from another body of mass 12 kg. The combination is being pulled by a string with an acceleration of 2.2 m $s^{- 2}$ . The tension $T_{1}$ and $T_{2}$ will be respectively : (use g = 9.8 $m / s^{2}$ )

200 N, 80 N

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220 N, 90 N

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240 N, 96 N

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260 N, 96 N

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Solution

The correct option is C 240 N, 96 N
Draw the separate free body diagram of both the block
so for 8kg block $T_{2}$ -mg=ma
hence $T_{2}$ =m $\times$ (g+a)=8 $\times$ (9.81=2.2)=96.08 N
for 12kg block $T_{1}$ -( $m_{1} + m_{2}$ )(g)=( $m_{1} + m_{2}$ )(a)
$T_{1}$ =( $8 + 12$ )( $9.81 + 2.2$ )=240.2 N

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A body of mass 8 kg is hanging from another body of mass 12 kg. The combination is being pulled by a string with an acceleration of 2.2 m s−2. The tension T1 and T2 will be respectively : (use g = 9.8 m/s2 )

The correct option is C 240 N, 96 NDraw the separate free body diagram of both the block so for 8kg block T2-mg=mahenceT2=m×(g+a)=8×(9.81=2.2)=96.08 Nfor 12kg block T1-(m1+m2)(g)=(m1+m2)(a)T1=(8+12)(9.81+2.2)=240.2 N

A body of mass 8 kg is hanging from another body of mass 12 kg. The combination is being pulled by a string with an acceleration of 2.2 m $s^{- 2}$ . The tension $T_{1}$ and $T_{2}$ will be respectively : (use g = 9.8 $m / s^{2}$ )

The correct option is C 240 N, 96 N
Draw the separate free body diagram of both the block
so for 8kg block $T_{2}$ -mg=ma
hence $T_{2}$ =m $\times$ (g+a)=8 $\times$ (9.81=2.2)=96.08 N
for 12kg block $T_{1}$ -( $m_{1} + m_{2}$ )(g)=( $m_{1} + m_{2}$ )(a)
$T_{1}$ =( $8 + 12$ )( $9.81 + 2.2$ )=240.2 N