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2 years ago

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A forklift raises a 1020 N crate 3.50 m up to a shelf l. How much work is done by the forklift on the crate? The forklift does J

work on the crate.

2 answers:

Rom4ik [11]2 years ago

7 0

A forklift raises a 1,020 N crate 3.50 m up to a shelf. How much work is done by the forklift on the crate?

The forklift does

⇒ 3,570 J of work on the crate. the answer is 3570

lord [1]2 years ago

5 0

Answer : Work done on the crate is 3570 J

Explanation :

Force acting on forklift due to its mass is 1020 N

Distance covered in lifting it, d = 3.5 m

Mathematically, the work done is defined as :

$W=F\times d$

So, $W=1020\ N\times 3.5\ m$

$W=3570\ J$

The forklift does 3570 J work on the crate.

Hence, this is the required solution.

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A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behav

gayaneshka [121]

Answer:

a) x = 0.098

b) x = 2.72 m

Explanation:

(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.

When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

Then, you have:

$K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2$ (1)

m: mass of the person = 62kg

k: spring constant = ?

v: velocity of the person just when he touches the fire net = ?

x: elongation of the fire net = 1.4 m

Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:

$v^2=v_o^2+2gy$

vo: initial velocity = 0 m/s

g: gravitational acceleration = 9.8 m/s^2

y: height from the person jumps = 20.0m

$v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}$

With this value you can find the spring constant k from the equation (1):

$mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}$

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

$W=F_e\\\\mg=kx$

you solve the last expression for x:

$x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m$

When the person is lying on the fire net the elongation of the fire net is 0.098m

b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:

$v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}$

Next, you calculate x from the equation (1):

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m$

The net fire is stretched 2.72 m

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2 years ago

What is the speed of a transverse wave in a rope of length 2 meters and mass 0.06

IgorLugansk [536]

Answer:

16.3 wave speed.

Explanation:

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3 years ago

Heating food under a heat lamp is an example of heat transfer by

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2 years ago

The speed of a box traveling on a horizontal friction surface changes from vi = 13 m/s to vf = 11.5 m/s in a distance of d = 8.5

KiRa [710]

Answer:

0.68 s

Explanation:

We are given that

Initial velocity of box= $u=13m/s$

Final velocity of box=v=11.5 m/s

Distance=d=8.5 m

We have to find the time taken by box to slow by this amount.

We know that

$v^2-u^2=2as$

Substitute the values

$(11.5)^2-(13)^2=2a(8.5)$

$132.25-169=17a$

$-36.75=17a$

$a=\frac{-36.75}{17}=-2.2m/s^2$

We know that

Acceleration= $a=\frac{v-u}{t}$

Substitute the values

$-2.2=\frac{11.5-13}{t}$

$-2.2=\frac{-1.5}{t}$

$t=\frac{1.5}{2.2}=0.68 s$

Hence, the time taken by box to slow by this amount=0.68 s

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3 years ago

When you are on a roller coaster, you are constantly transforming from Potential to Kinetic energy and back. Explain how these e

andreev551 [17]

Answer:

The two types of energy possessed by the roller coaster are:

- Potential energy: it is the energy possessed by the roller coaster due to its position. It is calculated as

$PE=mgh$

where

m is the mass of the roller coaster

g is the acceleration due to gravity

h is the height of the roller coaster relative to the ground

- KInetic energy: it is the energy possessed by the roller coaster due to its motion. It is calculated as

$KE=\frac{1}{2}mv^2$

where

v is the speed of the roller coaster

Moreover, according to the law of conservation of energy, the total mechanical energy of the roller coaster (the sum of potential+kinetic energy) is constant during the motion:

$E=PE+KE=const.$

This implies that:

- When PE increases (because h increases), KE decreases (because v decreases)

- When PE decreases (because h decreases), KE increases (because v increases)

Now we can apply these conclusions to the motion of the roller coaster:

- When it moves from A to B, potential energy is converted into kinetic energy, so PE decreases and KE increases

- When it moves from B to C, kinetic energy is converted into potential energy, so PE increases and KE decreases

- When it moves from C to D, potential energy is converted into kinetic energy, so PE decreases and KE increases

- When it moves from D to E, kinetic energy is converted into potential energy, so PE increases and KE decreases

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2 years ago