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3 years ago

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A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn doo

r, the cow walks from x = 0 to x = 6.9 m as you apply a force with x-component F _ { x } = - [ 20.0 N + ( 3.0 N / m ) x ]F x =−[20.0N+(3.0N/m)x]. How much work does the force you apply do on the cow during this displacement?

1 answer:

Tresset [83]3 years ago

7 0

Answer:

The work done is -209.42 J.

Explanation:

F(x) = (- 20 - 3 x ) N

x = 0 to x = 6.9 m

Here, the force is variable in nature, so the work done by the variable force is given by

$W =\int F dx\\\\W =\int_{0}^{6.9}(-20- 3x dx )\\\\W= \left [ - 20 x - 1.5 x^2 \right ]_{0}^{6.9}\\\\W = - 20 (6.9 - 0) - 1.5(6.9\times 6.9 - 0)\\\\W =- 138 - 71.42\\\\W = - 209.42 J$

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In the first part of the motion, the car accelerates at rate $a_1$ , so the final velocity after a time t is:

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Since it starts from rest,

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So the time taken for this part of the motion is

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$v_2 = v - a_2 t$

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Re-arranging the equation,

$t_2=\frac{v}{a_2}$ (2)

So the total time taken for the trip is

$t=\frac{v}{a_1}+\frac{v}{a_2}$

(b) $d=\frac{v^2}{2a_1}+\frac{v^2}{2a_2}$

In the first part of the motion, the distance travelled by the car is

$d_1 = u t_1 + \frac{1}{2}a_1 t_1^2$

Substituting u = 0 and $t_1=\frac{v}{a_1}$ (1), we find

$d_1 = \frac{1}{2}a_1 \frac{v^2}{a_1^2} = \frac{v^2}{2a_1}$

In the second part of the motion, the distance travelled is

$d_2 = v t_2 - \frac{1}{2}a_2 t_2^2$

Substituting $t_2=\frac{v}{a_2}$ (2), we find

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Explanation:

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A car of mass 1600 kg traveling at 27.0 m/s is at the foot of a hill that rises vertically 135 m after travelling a distance of

Zarrin [17]

Answer:

Neglecting any frictional losses, the average power delivered by the car's engine is 10565 W

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The energy at the bottom is only the Kinect energy (K_1) of the car in motion, but in the top, the energy is the sum of its Kinect energy (K_2), potential energy (P) and the work (W) done by the engine.

$K_1 = K_2 + P + W$

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$K=\frac{1}{2}mV^2\\P=mgh$

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$t=\frac{3200m}{20m/s}=160s$

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