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mario62 [17]
3 years ago
12

A child on a tricycle is moving at a speed of 1.40 m/s at the start of a 2.25 m high and 12.4 m long incline. The total mass is

48.0 kg, air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N, and the speed at the lower end of the incline is 6.50 m/s. Determine the work done (in J) by the child as the tricycle travels down the incline.
Physics
1 answer:
goblinko [34]3 years ago
5 0

Answer:

The work done by the child as the tricycle travels down the incline is 416.96 J

Explanation:

Given;

initial velocity of the child, v_i = 1.4 m/s

final velocity of the child, v_f = 6.5 m/s

initial height of the inclined plane, h = 2.25 m

length of the inclined plane, L = 12.4 m

total mass, m = 48 kg

frictional force, f_k = 41 N

The work done by the child is calculated as;

\Delta E_{mech} = W - f_{k} \Delta L\\\\W = \Delta E_{mech}  + f_{k} \Delta L\\\\W = (K.E_f - K.E_i) + (P.E_f - P.E_i) + f_{k} \Delta L\\\\W = \frac{1}{2} m(v_f^2 - v_i^2) + mg(h_f - h_i) + f_{k} \Delta L\\\\W = \frac{1}{2} \times 48(6.5^2 - 1.4^2) + 48\times 9.8(0-2.25) + (41\times 12.4)\\\\W = 966.96  \ - \ 1058.4 \ + \ 508.4\\\\W = 416.96 \ J

Therefore, the work done by the child as the tricycle travels down the incline is 416.96 J

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A concave mirror has a radius of curvature of 20 cm. What is it's focal length? If an object is placed 15 cm in front of it, where would the image be formed? What is it's magnification?

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Explanation:

Given:

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Radius of curvature of the mirror, as C = 20 cm

Object distance in-front of the mirror = 15 cm

a.

Focal length:

Focal length is half of the radius of curvature.

Focal length of the mirror =  \frac{C}{2} = 10 cm

According to the sign convention we will put the mirror on (0,0) point, of the Cartesian coordinate open towards the negative x-axis.

Object and the focal length are also on the negative x-axis where focal length and image distance will be negative numerically.

b.

We have to find the object distance:

Formula to be use:

⇒ \frac{1}{focal\ length}= \frac{1}{image\ distance} + \frac{1}{object\ distance}

⇒ Plugging the values.

⇒ \frac{1}{-10} =\frac{1}{image\ distance}+\frac{1}{-15}

⇒ \frac{1}{-10} -\frac{1}{-15}=\frac{1}{image\ distance}

⇒ \frac{1}{-10} + \frac{1}{15}=\frac{1}{image\ distance}

⇒ \frac{-3+2}{30} =\frac{1}{image\ distance}

⇒ \frac{-1}{30} =\frac{1}{image\ distance}

⇒ -30\ cm=image\ distance

Image will be formed towards negative x-axis 30 cm away from the pole.

c.

Magnification (m) is the negative ratio of mage distance and object distance:

⇒ m=-\frac{image\ distance}{object\ distance}

⇒ m=-\frac{(-30)}{(-15)}

⇒ m=-2

The focal length of the concave mirror, is of 10 cm, object distance is 30 cm and magnification is -2.

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