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A heavy block of mass 625 kg is suspended by a cable is moving downwards with an initialvelocity of 7.3 m/s. If the tension in the cable as the block comes to rest is 5000 Ncalculate the distance travelled by the block(a) 25.0 m(b) 13.6 m(c) 2.72 m(d) 14.8 m

Question
A heavy block of mass 625 kg is suspended by a cable is moving downwards with an initial
velocity of 7.3 m/s. If the tension in the cable as the block comes to rest is 5000 N
calculate the distance travelled by the block
(a) 25.0 m
(b) 13.6 m
(c) 2.72 m
(d) 14.8 m
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A heavy block of mass 625 kg is suspended by a cable is moving downwards with an initial velocity of 7.3 m/s. If the tension in the cable as the block comes to rest is 5000 N calculate the distance travelled by the block (a) 25.0 m (b) 13.6 m (c) 2.72 m (d) 14.8 m

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Step 1

Net force is,

mg-T ma
Substituting the values,
(625 kg 9.8m/s2)-s000 N = ( 625 kg )a
1125N
625kg
-1.8m/s2
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mg-T ma Substituting the values, (625 kg 9.8m/s2)-s000 N = ( 625 kg )a 1125N 625kg -1.8m/s2

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Step 2

Acceleration is,

-
At
Substituting the values,
1.8m/s73m/s- Om/s
7.3m/s
t=
1.8m/s
= 4.1s
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- At Substituting the values, 1.8m/s73m/s- Om/s 7.3m/s t= 1.8m/s = 4.1s

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Step 3

Distance travelled ...

-
S=
2a
Substituting the values,
. (7.3m/s)-(0m/s)
S =
2(1.8m/s
14.8m
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- S= 2a Substituting the values, . (7.3m/s)-(0m/s) S = 2(1.8m/s 14.8m

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