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2 years ago

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A person pushes a 15.7-kg shopping cart at a constant velocity for a distance of 25.9 m on a flat horizontal surface. She pushes

in a direction 23.7 ° below the horizontal. A 32.7-N frictional force opposes the motion of the cart. (a) What is the magnitude of the force that the shopper exerts? Determine the work done by (b) the pushing force, (c) the frictional force, and (d) the gravitational force.

1 answer:

valentinak56 [21]2 years ago

6 0

Answer:

a) F = 35.7 N, b) W = 846.7 J, c) W = - 846.9 J, d) W=0

Explanation:

a) For this exercise let's use Newton's second law, let's set a reference frame with the x-axis horizontally

let's break down the pushing force.

cos (-23,7) = Fₓ / F

sin (-237) = F_y / F

Fₓ = F cos 23.7 = F 0.916

F_y = F sin (-23.7) = - F 0.402

Y axis

N- W - F_y = 0

N = W + F 0.402

X axis

Fₓ - fr = 0

F 0.916 = fr

F = fr / 0.916

F = 32.7 / 0.916

F = 35.7 N

It is asked to calculate several jobs

b) the work of the pushing force

W = fx x

W = 35.7 cos 23.7 25.9

W = 846.7 J

c) friction force work

W = F x cos tea

friction force opposes movement

W = - fr x

W = - 32.7 25.9

W = - 846.9 J

d) The work of the force would gravitate, as the displacement and the force of gravity are at 90º, the work is zero

W = 0

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At one point in a pipeline, the water's speed is 3.57 m/s and the gauge pressure is 68.7 kPa. Find the gauge pressure at a secon

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The pressure at point 2 is $P_2 = 254.01 kPa$

Explanation:

From the question we are told that

The speed at point 1 is $v_1 = 3.57 \ m/s$

The gauge pressure at point 1 is $P_1 = 68.7kPa = 68.7*10^{3}\ Pa$

The density of water is $\rho = 1000 \ kg/m^3$

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Let the diameter at point 1 be $d_1$ then the diameter at point two will be

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Here $A_1 , A_2$ are the area at point 1 and 2

Now given that the are is directly proportional to the square of the diameter [i.e $A= \frac{\pi d^2}{4}$ ]

which can represent as

$A \ \ \alpha \ \ d^2$

=> $A = c d^2$

where c is a constant

so $\frac{A_1}{d_1^2} = \frac{A_2}{d_2^2}$

=> $\frac{A_1}{d_1^2} = \frac{A_2}{4d_1^2}$

=> $A_2 = 4 A_1$

Now from the continuity equation

$A_1 v_1 = 4 A_1 v_2$

=> $v_2 = \frac{v_1}{4}$

=> $v_2 = \frac{3.57}{4}$

$v_2 = 0.893 \ m/s$

Generally the Bernoulli equation is mathematically represented as

$P_1 + \frac{1}{2} \rho v_1^2 + \rho * g * h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho * g * h_2$

So

$P_2 = \rho * g (h_1 -h_2 )+P_1 + \frac{1}{2} * \rho (v_1^2 -v_2 ^2 )$

=> $P_2 = \rho * g (h_1 -(h_1 -18.3) + P_1 + \frac{1}{2} * \rho (v_1^2 -v_2 ^2 )$

substituting values

$P_2 = 1000 * 9.8 (18.3) )+ 68.7*10^{3} + \frac{1}{2} * 1000 ((3.57)^2 -0.893 ^2 )$

$P_2 = 254.01 kPa$

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