A body of mass 450g changes it speed from 5ms¹ to 25ms¹. what is the work done by the body?

A book prone to air resistance is released from rest 300 m

yaroslaw [1]

Answer:

Approximately $73\%$ .

(Assuming that $g = \rm 9.81\; m \cdot s^{-2}$ .)

Explanation:

The mechanical energy of an object is the sum of its potential energy and its kinetic energy. It will be shown that the exact mass of this object doesn't matter. For ease of calculation, let $m(\text{book})$ represent the mass of the book.

The initial potential energy of the book is

$\begin{aligned}U(300\; \text{m}) &= m(\text{book}) \cdot g \cdot \Delta h + U(0\; \text{m}) \cr &=(9.81 \times 300) \cdot m(\text{book})\cr &= \left(2.943\times 10^3\right) \cdot m(\text{book})\end{aligned}$ .

The book was initially at rest when it was released. Hence, its initial kinetic energy would be zero. Hence, the initial mechanical energy of the book-Earth system would be $(2.943\times 10^3) \cdot m(\text{book})$ .

When the book was about to hit the ground, its speed is $\rm 40\; m \cdot s^{-1}$ . Its kinetic energy would be:

$\begin{aligned} \text{KE} &= \frac{1}{2} \, m(\text{book}) \cdot v^{2} \cr &= \left(\frac{1}{2} \times 40^2\right)\cdot m(\text{book}) \cr &= \left(8.00\times 10^2\right)\cdot m(\text{book})\end{aligned}$ .

The question implies that the potential energy of the book near the ground is zero. Hence, the mechanical energy of the system would be $\left(8.00\times 10^2\right)\cdot m(\text{book})$ when the book was about to hit the ground.

The amount of mechanical energy lost in this process would be equal to:

$\begin{aligned}&\left(2.943\times 10^3\right) \cdot m(\text{book}) - \left(8.00\times 10^2\right)\cdot m(\text{book}) \cr &=\left(2.143\times 10^3\right)\cdot m(\text{book})\end{aligned}$ .

Divide that with the initial mechanical energy of the system to find the percentage change. Note how the mass of the book, $m(\text{book})$ , was eliminated in this process.

$\begin{aligned}&\frac{\left(2.143\times 10^3\right)\cdot m(\text{book})}{\left(2.943\times 10^3\right) \cdot m(\text{book})}\times 100\% \cr &= \frac{2.143\times 10^3}{2.943\times 10^3}\times 100\% \cr & \approx 73\%\end{aligned}$ .

5 0

3 years ago

Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 64.7 N, Jill pulls with 61.5 N in the

USPshnik [31]

Answer:

F = 241.8 N θ = -18.7

Explanation:

To find the net force, we use Newton's second law, for this we decompose the force using trigonomer

Force 2

F2 = 61.5 N

cos 45 = F₂ₓ / F2 F₂ₓ = F2 cos 45

sin 45 = F₂_y / F2 F₂_y = F2 sin 45

force 3

F3 = 171 N

cos 45 = F₃ₓ / F3

sin 45 = F₃_y / F3

the total force can be found with its components

axis x direction (East-West)

Fₓ = 64.7 +61.5 cos 45 + 171 cos 45

Fₓ = 229.1 N

Y axis (direction North -Sur)

F_y = 61.5 sin 45 - 171 sin 45

F_y = - 77,428 N

the resulting force is

F = Fx i ^ + Fy j ^

F = (229.1 i⁻ 77,428 j⁾ N

we can use the Pythagorean theorem to find the module

F = √ (229.1 2 + 77,428 2)

F = 241.8 N

let's use trigonometry to find the direction

tan θ = F_y / Fₓ

θ = tan⁻¹ (F_y / Fₓ)

θ = tan⁻¹ (-77,428 / 229.1)

θ = -18.7

4 0

3 years ago

If a tube is placed vertically into a container of water, and the water level inside the tube is the same as the water level on

dybincka [34]

It's equal to d external atmospheric pressure

6 0

3 years ago

if an object is thrown directly upwards remains in the air for 5.6s before it returns to its original position, how long did it

Mila [183]

Okay so an object gets thrown upwards so the half way point of the trip would be the maximum height (before it starts coming back down). if the object stays in the air for a total of 5.6s, then that is the time it takes to go up and come back down. to find the time to maximum height, half the time of the whole trip

5.6s/2 = 2.8s

6 0

4 years ago

Read 2 more answers

Two 75 W (120 V) lightbulbs are wired in series, then the combination is connected to a 120 V supply. Part A How much power is d

Elenna [48]

Answer:

300 W

Explanation:

power of each bulb P = 75 W

voltage in the circuit = 120 V

we know that electrical power P = IV ....1

and V = IR