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1. I'd be grateful if someone could help me with this question. I thought you would use F=ma, but i'm wrong. Thanks

A lift and its passengers with a total mass of 500kg accelerates upwards at 2ms^-2. Assume that g=10ms^-2. What is the tension in the cable holding the lift?

Thanks.
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3. (Original post by liamsullivan)
I'd be grateful if someone could help me with this question. I thought you would use F=ma, but i'm wrong. Thanks

A lift and its passengers with a total mass of 500kg accelerates upwards at 2ms^-2. Assume that g=10ms^-2. What is the tension in the cable holding the lift?

Thanks.
Well F=ma is useful here, but you've got to remember that the tension in the cable isn't zero when the lift is stationary.
4. (Original post by liamsullivan)
I'd be grateful if someone could help me with this question. I thought you would use F=ma, but i'm wrong. Thanks

A lift and its passengers with a total mass of 500kg accelerates upwards at 2ms^-2. Assume that g=10ms^-2. What is the tension in the cable holding the lift?

Thanks.
You need to consider the forces acting on the elevator to provide for this accelration. F=ma is what you use but F is a number of different forces (making up the net force) acting on the elevator not just one. There is a tension upwards and the weight of the passengers + lift downwards. Since accerlation upwards then the tension must be greater as tension is up and weight force is down. So:

T - mg = ma
T - 500g = (500x2)
T = 5900N

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