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The velocity ...

Question

The velocity time graph of a lift moving upwards has been shown below. Let $T_{1}, T_{2}$ and $T_{3}$ be the tensions in the elevator cable during the three time intervals $Δ t_{1}, Δ t_{2}$ and $Δ t_{3}$ , then $T_{1} : T_{2} : T_{3}$

11 : 10 : 9

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19 : 10 : 11

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11 : 10 : 12

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11 : 10 : 8

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Solution

The correct option is D $11 : 10 : 8$
$T = m g + m a$
$T_{1} = m (10 + 1) = 11 m$ $(∵ a = 1 {m/s}^{2}, 0 - 10 s)$
$T_{2} = m (10 + 0) = 10 m$ $(∵ a = 0, 10 - 20 s)$
$T_{3} = m (10 - 2) = 8 m$ $(∵ a = - 2, 20 - 25 s)$
$∴ T_{1} : T_{2} : T_{3} = 11 : 10 : 8$

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The velocity time graph of a lift moving upwards has been shown below. Let T1,T2 and T3 be the tensions in the elevator cable during the three time intervals Δt1,Δt2 and Δt3, then T1:T2:T3

The correct option is D 11:10:8T=mg+ma T1=m(10+1)=11m (∵a=1 m/s2,0−10 s) T2=m(10+0)=10m (∵a=0,10−20 s) T3=m(10−2)=8m (∵a=−2,20−25 s) ∴T1:T2:T3=11:10:8

The velocity time graph of a lift moving upwards has been shown below. Let $T_{1}, T_{2}$ and $T_{3}$ be the tensions in the elevator cable during the three time intervals $Δ t_{1}, Δ t_{2}$ and $Δ t_{3}$ , then $T_{1} : T_{2} : T_{3}$

The correct option is D $11 : 10 : 8$
$T = m g + m a$
$T_{1} = m (10 + 1) = 11 m$ $(∵ a = 1 {m/s}^{2}, 0 - 10 s)$
$T_{2} = m (10 + 0) = 10 m$ $(∵ a = 0, 10 - 20 s)$
$T_{3} = m (10 - 2) = 8 m$ $(∵ a = - 2, 20 - 25 s)$
$∴ T_{1} : T_{2} : T_{3} = 11 : 10 : 8$