Assuming that the box is moving when it is being pulled, Work is done on the box.

So work is the Force times the distance

W=Fd

But what is work actually ? When something moves due to force over some change in distance, it have energy.

But where does this energy come from ? Does it magically appear ? The energy comes from the applied force onto the box.

So the energy have been transferred. And it’s like that throughout the universe

Now to save time, I’ll just tell you the answer: kinetic energy

:)

3 0

3 years ago

Q1) After applying a force of 1000 N an object of mass 2000 kg will achieve what acceleration?

wlad13 [49]

Answer:

F=mass x acceleration

1000= 2000 x a

a=1000/2000

a=1/2 m/sec^2

mark it as brainliest!!!

7 0

3 years ago

Read 2 more answers

A 0.305 kg book rests at an angle against one side of a bookshelf. The magnitude and direction of the total force exerted on the

tankabanditka [31]

Answer

given,

$F_L= 1.52\ N$

$\theta_L= 31^0$

mass of book = 0.305 Kg

so, from the diagram attached below

$F_L cos {\theta_L} + F_b sin {\theta_b} = m g$

$1.52 times cos {31^0} + F_b sin {\theta_b} = 0.305 \times 9.8$

$F_b sin {\theta_b} = 2.989 -1.303$

$F_b sin {\theta_b} = 1.686$

computing horizontal component

$F_b cos {\theta_b} = F_L sin {\theta_L}$

$cos {\theta_b} = \dfrac{F_L sin {\theta_L}}{F_b}$

$cos {\theta_b} = \dfrac{1.52 \times sin {31^0}}{1.686}$

$cos {\theta_b} = 0.464$

θ = 62.35°

5 0

3 years ago

A 30-kg shopping cart full of groceries sitting at the top of

Evgesh-ka [11]

Answer:

0.0102 m or 1 cm

Explanation:

Let g = 10m/s2

The potential energy of the shopping cart of the top of the hill is:

$E_p = mgh = 30*9.8*2 = 600 J$

When the cart gets to the bottom of the hill, all this potential energy is converted to kinetic energy:

$E_k = mv^2/2 = 600 J$

$v^2 = \frac{600*2}{30} = 40$

$v = \sqrt{39.2} = 6.324 m/s$

As the cart stop due to the stump, the can of peaches flies with the same speed.

By Newton's 3rd law, the car would exert a 490N force on the can too

The deceleration of the can would then be:

$a = F/m = 490/0.25 = 1960 m/s^2$

This force would stop the can, but not without making a dent, aka a traveled distance on the car skin

We can use the following equation of motion to find out the distance traveled by the can:

$v^2 - v_0^2 = 2a\Delta s$

where v = 0 m/s is the final velocity of the can when it stops, $v_0^2$ = 40m/s is the initial velocity of the can when it hits, a = -1960 m/s2 is the deceleration of the can, and $\Delta s$ is the distance traveled, which we care looking for:

$0 - 40 = 2*(-1960)*\Delta s$

$\Delta s = \frac{40}{2*1960} = 0.0102 m$ or 1 cm

3 0

3 years ago