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Aleksandr-060686 [28]

2 years ago

12

Adam pushes a 47 kg boat across a frozen lake. The coefficient of kinetic friction between the boat and the ice is 0.13 and the

boat acceleration at 0.6 m/s^2. Calculate the magnitude of Adams force in the boat.

1 answer:

yuradex [85]2 years ago

5 0

Answer:

The magnitude of Adams' force is 88 N

Explanation:

According to Newton's law ∑ forces in direction of motion is equal to

mass multiplied by the acceleration

There are two forces here the friction force and Adams' force

Adams' force is in direction of motion

Friction force is opposite to direction of motion

→ Friction force = μ R,

where μ is the coefficient of friction and R is the normal reaction force

of the boat

→ R = mg

where m is the mass of the boat and g is the acceleration of gravity

→ m = 47 kg , g = 9.8 m/s²

→ R = mg

Substitute the values in the rule

→ R = 47 × 9.8 = 460.6 N

→ Friction force = μ R

→ μ = 0.13 , R = 460.0 N

Substitute the values in the rule

→ Friction force = 0.13 × 460.0 = 59.878 N

→ ∑ Forces in direction of motion = mass × acceleration

→ F - Friction force = mass × acceleration

→ Friction force = 59.878 N , m = 47 kg , a = 0.6 m/s²

Substitute the values in the rule

→ F - 59.878 = 47 × 0.6

→ F - 59.878 = 28.2

Add 59.878 to both sides

→ F = 88.078 N ≅ 88 N

<em>The magnitude of Adams' force is 88 N </em>

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Answer:

$W=17085KJ$

Explanation:

From the question we are told that:

Height $H=16m$

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Density of water $\rho=1000kg/m^3$

Generally the equation for Volume of water is mathematically given by

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$dv=\frac{\piR^2}{H^2}(H-y)^2dy$

Where

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Generally the equation for Work done to pump water is mathematically given by

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Substituting dv

$dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)$

$dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy$

Therefore

$W=\int dw$

$W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy$

$W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)$

$W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0$

$W=3420.84*0.25[2401-65536]$

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Lapatulllka [165]

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