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2 years ago

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A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.50m/s . The pulling force is 100 N parallel to t

he incline, which makes an angle of 20.0° with the horizontal. The coefficient of kinetic friction is 0.400 , and the crate is pulled 5.00m .(a) How much work is done by the gravitational force on the crate?

1 answer:

sergey [27]2 years ago

5 0

167.67 Joules of work are done by the gravitational force on the crate.

Force is defined as the product of the mass of the object and acceleration. The SI unit of the force of Newton.

Gravitational force is the force applied by gravity on an object. The gravity of the earth is 9.8 m/s².

Force = Mass × Acceleration

F = m × a

Mass of the crate = 10 kg

The initial speed of the crate = 1.50 m/s

The pulling force = 100 N

Horizontal angle made by the crate = 20 °

Coefficient of kinetic friction = 0.400

The crate is pulled to a height of 5 m.

$Sin θ= \frac{height}{distance}$

$= \frac{h}{d}$

$h = dSin θ$

Work done by the gravitational force on the crate is,

Work done by gravitational force = mass × gravity × height

$W_{g} = mgh$

$W_{g} = m \times g \times d \times sinθ$

$= 10 \times 9.8 \times 5 \times sin20°$

$= 167.67 \: J$

Therefore, 167.67 Joules of work is done by the gravitational force on the crate.

To know more about force, refer to the below link:

brainly.com/question/13191643

#SPJ4

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Answer:

Average speed=1.5 m/s

Frequency of pendulum=93.75Hz

Explanation:

We are given

Frequency, $f=0.100Hz$

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Wavelength, $\lambda'=0.32m$

We have to find the average speed and frequency of pendulum.

We know that

Speed, $v=\lambda f$

Using the formula

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Hence, the average speed =1.5m/s

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Using the formula

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Hence, the frequency of a pendulum=93.75Hz

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3 years ago

Four charges 7 × 10−9 C at (0 m, 0 m), −9 × 10−9 C at (3 m, 3 m), 7 × 10−9 C at (1 m, 3 m), and −8 × 10−9 C at (−3 m, 2 m), are

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Answer:

Magnitude of the resulting force on the 7 nC charge at the origin:

Fn₁= 23.95*10⁻⁹ N

Explanation:

Look at the attached graphic:

Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.

q₁ experiences three forces (F₂₁,F₃₁,F₄₁) and we calculate them with Coulomb's law:

F = (k*q₁*q)/(d)²

$d_{12} = \sqrt{3^{2}+3^{2} } = \sqrt{18}$ m : distance from q₁ to q₂

(d₁₂)² = 18 m²

$d_{13} =\sqrt{1^{2}+3^{2} } = \sqrt{10}$ m : distance from q₁ to q₃

(d₁₃)² = 10 m²

$d_{14} =\sqrt{3^{2}+2^{2} } = \sqrt{13}$ m : distance from q₁ to q₄

(d₁₄)² = 13 m²

K= 8.98755 × 10⁹ N *m²/C²

q₁= 7*10⁻⁹C

k*q₁=8.98755*10⁹ *7*10⁻⁹= 62.9

F₂₁= (62.9)*(9* 10⁻⁹) /(18) = 31.45*10⁻⁹ C

F₃₁= (62.9)*(7* 10⁻⁹) /(10) = 44*10⁻⁹ C

F₄₁= (62.9)*(8* 10⁻⁹) /(13) = 38.7*10⁻⁹ C

x-y components of the net force on q₁ (Fn₁):

α= tan⁻¹(3/3)= 45° , β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°

Fn₁x = F₂₁x+ F₃₁x+F₄₁x

F₂₁x =+ F₂₁*cosα =+ (31.45*10⁻⁹)* (cos 45°) = +22.24 *10⁻⁹ N

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F₄₁x= -F₄₁*cosθ = -(38.7*10⁻⁹)* (cos 33.69°) = -32.2*10⁻⁹ N

Fn₁x = (+22.24 - 13.91 - 32.2)*10⁻⁹ N

Fn₁x = -23.87 *10⁻⁹ N

Fn₁y = F₂₁y+ F₃₁y+F₄₁y

F₂₁x =+ F₂₁*sinα =+ (31.45*10⁻⁹)* (sin 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*sinβ = - ( 44*10⁻⁹)* (sin 71.56°) = -41.74 *10⁻⁹ N

F₄₁x= +F₄₁*sinθ = +(38.7*10⁻⁹)* (sin 33.69°) =+21.47*10⁻⁹ N

Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N

Fn₁y = 1.97*10⁻⁹ N

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$F_{n1} =\sqrt{(Fn_{1x} )^{2}+(Fn_{1y} )^{2} }$

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