Physics question （urgent）

2013-05-20 6:32 am

a uniform chain with a mass of 0.33kg and a length of 0.25m is coiled on a table. One end is pulled upward at a constant rate of 0.47m/s.
(a) calculate the net force acting on the chain
(b) at the instant when 0.15m of the chain has been lifted off the table, how much force must be applied to the end being raised.

I want the steps and answers. Please!!!

回答 (1)

天同

2013-05-20 3:41 pm

✔ 最佳答案

(a) Linear density of chain p = 0.33/0.25 kg/m = 1.32 kg/m

Since net force Fn equals to the rate of change of momentum,
i.e. Fn = d(mv)/dt
where m is the mass of chain being in the air and v is the speed of the lifting chain.

thus, Fn = v(dm/dt) [ because v being constant ]
But m = p.L where L is the length of chain in the air
dm/dt = p.(dL/dt) = p.v
Hence, Fn = v.(pv) = pv^2 = 1.32 x 0.47^2 N = 0.2916 N

(b) Let F be the required force
Fn = F - mg where g is the acceleration due to gravity
Hence, F = Fn + mg = Fn + pLg = (0.2916 + 1.32 x 0.15g) N
F = 2.27 N (take g = 10 m/s^2)

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