Ask question

Ask question

All categories

2 years ago

7

If a boy lifts a mass of 6kg to a height of 10m and travels horizontally with a constant velocity of 4.2m/s, calculate the work

done? Explain your answer.

1 answer:

kirza4 [7]2 years ago

4 0

Answer:

W = 641.52 J

Explanation:

The work done here will be the sum of potential energy and the kinetic energy of the boy. Here potential energy accounts for vertical motion part while the kinetic energy accounts for the horizontal motion part:

$Work\ Done = Kinetic\ Energy + Potential\ Energy\\\\W = K.E +P.E\\\\W = \frac{1}{2}mv^2+mgh\\\\$

where,

W = Work Done = ?

m = mass = 6 kg

v = speed = 4.2 m/s

g = acceleration dueto gravity = 9.81 m/s²

h = height = 10 m

Therefore,

$W = \frac{1}{2}(6\ kg)(4.2\ m/s)^2+(6\ kg)(9.81\ m/s^2)(10\ m)$

W = 52.92 J + 588.6 J

<u>W = 641.52 J</u>

You might be interested in

2. How can people know that they've learned?

photoshop1234 [79]

Answer: People know that they jap e learned by their mistakes or by someone ya being them something

4 0

3 years ago

You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 77.2 kg hop on board for a ride

lions [1.4K]

To solve this problem it is necessary to apply the kinematic equations of motion and Hook's law.

By Hook's law we know that force is defined as,

$F= kx$

Where,

k = spring constant

x = Displacement change

PART A) For the case of the spring constant we can use the above equation and clear k so that

$k= \frac{F}{x}$

$k = \frac{mg}{x}$

$k= \frac{77.2*9.8}{0.0637}$

$k = 11876.92N/m$

Therefore the spring constant for each one is 11876.92/2 = 5933.46N/m

PART B) In the case of speed we can obtain it through the period, which is given by

$T = \frac{2\pi}{\omega}$

Re-arrange to find \omega,

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{2.14}$

$\omega = 2.93rad/s$

Then through angular kinematic equations where angular velocity is given as a function of mass and spring constant we have to

$\omega^2 = \frac{k}{m}$

$m = \frac{k}{\omega^2}$

$m = \frac{ 11876.92}{2.93}$

$m = 4093.55Kg$

Therefore the mass of the trailer is 4093.55Kg

PART C) The frequency by definition is inversely to the period therefore

$f = \frac{1}{T}$

$f = \frac{1}{2.14}$

$f = 0.4672 Hz$

Therefore the frequency of the oscillation is 0.4672 Hz

PART D) The time it takes to make the route 10 times would be 10 times the period, that is

$t_T = 10*T$

$t_T = 10 *2.14s$

$t_T = 21.4s$

Therefore the total time it takes for the trailer to bounce up and down 10 times is 21.4s

5 0

3 years ago

An object is held at rest and allowed to drop How far does the object fall in 3.3 seconds?

IgorC [24]

After x seconds, an object will fall $\frac{1}{2}at^2$ where a is acceleration due to gravity and t is time

so when t=3.3
the distance it will fall is $(\frac{1}{2})(9.8\frac{m}{s^2})(3.3m)^2$ =53.361m
it will fal 53.361 meters

6 0

3 years ago

an electric current of 285.0 ma flows 674. milliseconds. Calculate the amount of electric charge transported. Be sure your answe

yulyashka [42]

Answer:

<em>The amount of electric charge transported = 0.192 C</em>

Explanation:

Electric Charge: This is defined as the product of electric current and time in an electric circuit, The S.I unit of electric charge is Coulombs (C)

Q = It..................... Equation 1

Where Q = Electric charge, I = electric current, t = time.

<em>Given:</em> I = 285 mA, t = 674 milliseconds.

<em>Conversion: (i) Convert from 285 mA to A = (285/1000) A = 0.285 A</em>

<em> (ii) convert from 674 milliseconds to seconds = (674/1000) s = 0.674 s </em>

Substituting these values into equation 1

Q = 0.285 × 0.674

<em>Q = 0.192 C</em>

<em>Therefore the amount of electric charge transported = 0.192 C</em>

<em></em>

<em></em>

5 0

2 years ago

Where is a water or a mechanical wave in which the particles in the medium move in a direction perpendicular to the direction of

horrorfan [7]

transverse waves. ripples on water surface

6 0

3 years ago

Read 2 more answers