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VashaNatasha [74]

3 years ago

6

A net force of 3000 N is accelerating a 1200 kg elevator upward. If the elevator starts from rest, how long will it take to trav

el up 15 m?

1 answer:

alexdok [17]3 years ago

5 0

F = net force acting on the elevator in upward direction = 3000 N

m = mass of the elevator = 1200 kg

a = acceleration of the elevator = ?

Acceleration of the elevator is given as

a = F/m

a = 3000/1200

a = 2.5 m/s²

v₀ = initial velocity of the elevator = 0 m/s

Y = displacement of the elevator = 15 m

t = time taken

Using the kinematics equation

Y = v₀ t + (0.5) a t²

15 = (0) t + (0.5) (2.5) t²

t = 3.5 sec

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The gas is compressed and The temperature of the gas remains constant during this process.

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Since the gas is compressed ; then $v_f< v_i$

However;

$W =- nRT \ In \dfrac{V_f}{V_i}$

$W =- P_1V_1 \ In \dfrac{V_f}{V_i}$

The initial volume for the cylinder is calculated as ;

$v_1 = \pi r^2 h \\ \\ v_1 = \pi r^2 L \\ \\ v_1 = 3.14*(6*10^{-2})^2*(20*10^{-2}) \\ \\ v_1 = 2.261*10^{-3} \ m^3$

Replacing over values into the above equation; we have :

$100 = - ( 100*10^3 *2.261*10^{_3}) In (\dfrac{v_f}{v_i}) \\ \\ - In (\dfrac{v_f}{v_i})= \dfrac{100}{(100*10^3*2.261*10^{-3})} \\ \\ - In \ v_f + In \ v_i = \dfrac{100}{226.1} \\ \\ - In \ v_f = - In \ v_i + \dfrac{100}{226.1} \\ \\ - In \ v_f = - In (2.261*10^{-3} + \dfrac{100}{226.1 } \\ \\ - In \ v_f = 6.1 + 0.44 \\ \\ - In \ v_f = 6.54 \\ \\ - In \ v_f = -6.54 \\ \\ v_f = e^{-6.54} \\ \\ \mathbf{v_f = 1.445*10^{-3} m^3}$

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