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soldi70 [24.7K]

2 years ago

5

A force in the +x-direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 8.40 kg box that is sitting on the horizontal,

frictionless surface of a frozen lake. F(x) is the only horizontal force on the box. So, If the box is initially at rest at x=0, what is its speed after it has traveled 16.0 m? Express your answer to three significant figures and include the appropriate units.

1 answer:

Fynjy0 [20]2 years ago

6 0

Answer:v=7.24 m/s

Explanation:

Given

F(x)=18-0.530 x

mass m=8.40 kg

box is at rest at x=0

Distance traveled=16 m

$mv\frac{\mathrm{d} v}{\mathrm{d} x}=F=\left ( 18-0530x\right )dx$

$\int_{0}^{v}vdv=\int_{0}^{16}\left ( 18-0.530x\right )dx$

$\frac{mv^2}{2}=\left ( 18x-0530\frac{x^2}{2}\right )_0^{16}$

$v^2=\frac{288-67.841\times 2}{8.40}$

$v=7.24 m/s$

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A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

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Explanation:

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The charge on the second sphere is $q_2 = 8 \mu C = 8*10^{-6} \ C$

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$U = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.8 }$

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$Q = 0.3 + 0.18$

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Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.4 \ m$ is mathematically represented

$Q_f = KE_f + U_f$

Here $KE_f$ is the kinetic energy which is mathematically represented as

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substituting value

$KE_f = \frac{1 }{2} * ( 1.50 *10^{-3}) (v_2 )^2$

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And $U_f$ is the potential energy which is mathematically represented as

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substituting values

$U_f = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.4 }$

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From the law of energy conservation

$Q = Q_f$

So

$0.48 = 0.36 +(7.50 *10^{-4} v_2^2)$

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