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# Physics help please? ?

A rope is tied to a 0.50.5kg block and looped over a massless frictionless pulley. Consider the following situations: (A) A constant downward force of 9.89.8N is applied to the other end of the rope. (Specifically, the rope end is pulled so that the tension in the rope is 9.8N). (B) A 11kg mass is hanging off the other end of the rope. (C) A 11kg mass is attached to the other end of the rope and is resting on a frictionless horizontal surface. Rank the tensions in the rope for these three situations.

### 2 Answers

- MangalLv 42 months ago
(A) Tension, Ta = 9.8 N (given) = (1 kg)g

................................................. [g = 9.8 m/s², accel'n due to gravity]

(B) It is obvious that 1 kg mass will go down and 0.5 kg mass will go up.

Because 1 kg mass is going down

, ... ... ... Tb < (1 kg)g

Also, because 0.5 kg mass is going up,

... ... ... Tb > (0.5 kg)g

Thus, (1 kg)g > Tb > (0.5 kg)g

(C) It is obvious that 0.5 kg mass will go down. Hence

... ... ... Tc < (0.5 kg)g

Combining results we have obtained and the data given, we conclude

... ... ... Ta > Tb > Tc

[No arithmetic calculation required]

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- Anonymous2 months ago
Ta is 9.8 N

Tc = 0.5 kg * g

Net downward acceleration of the 1 kg mass (and upward acceleration of the 0.5 kg mass) = 1 kg * 9.81 m/sec^2 / (1 kg + 0.5 kg) = 6.54 m/sec^2. Since the 1 kg mass would accelerate downwards under gravity at 9.81 m/sec^2 if it was in free fall, Tb is pulling upwards slowing it down by ( 9.81 - 6.54 ) m/sec^2 * 1 kg = 3.27 N. This is Tb.

- MangalLv 42 months agoReport
Tc will be less than (0.5 kg)g

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