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3 years ago

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An inclined plane of effective length 4.5m is used to raise a load of 500N through a height of 1.5m .If the list is raise by for

ce of 200N
calculate frictional force between load and the plane

1 answer:

IrinaK [193]3 years ago

8 0

Answer:

1850 N

Explanation:

The formula for friction force between the load and plane is given as ;

F= μ*N

N = mg cos θ

To find θ, which is the angle the inclined plane makes with the ground at the height of 1.5 m

Sin θ = 1.5/4.5

Sin θ = 0.3333

Sin⁻{0.3333} = 19.50°

θ = 19.50°

Finding N , where m= 500 N , and g= 9.81

N = mg cos θ

N= 500 * 9.81 * cos 19.50°

N= 4624 N

Coefficient of kinetic friction is calculated as;

μ=F/W

μ = 200/500 = 0.4

The magnitude of kinetic friction is given as;

Fk= μ * N

Fk = 0.4 * 4624

Fk= 1850 N

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Answer:

(A). The current in the circuit is 19.25 mA.

(B). The store energy in the inductor is 7.04 μJ.

Explanation:

Given that,

Voltage = 8.2 V

Inductor = 38 mH

Resistance = 150 Ω

Time t = 0.110 ms

The battery has negligible internal resistance, so that the total resistance in the circuit is 150 ohms. Then use this equation for current at time t in terms of inductance

We need to calculate the current

Using formula of current

$I(t)=\dfrac{V}{R}\times(1-e^{-t\times\dfrac{R}{L}})$

Put the value into the formula

$I(t)=\dfrac{8.2}{150}\times(1-e^{-0.110\times10^{-3}\times\dfrac{150}{38\times10^{-3}}})$

$I(t)=0.01925\ A$

$I(t) = 19.25\ mA$

(B). We need to calculate the store energy in the inductor

Using formula of energy

$E=\dfrac{1}{2}LI^2$

Put the value into the formula

$E=\dfrac{1}{2}\times38\times10^{-3}\times(0.01925)^2$

$E=7.04\times10^{-6}\ J$

{tex]E=7.04\ \mu J[/tex]

Hence, (A). The current in the circuit is 19.25 mA.

(B). The store energy in the inductor is 7.04 μJ.

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3 years ago