A construction foreman exerts 1300 Newtons of force trying to move a 1200-kg block of concrete. How many Joules of work does he perform?

Answers

Answer 1

Answer: He does no work until the block starts to move ... an unlikely event.

If it ever does move, then the work he does is

(1300) x (the distance the block moves, in meters) .

The unit is 'joules.

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A nearsighted person has a far point of 40cm. What power spectacle lens is needed if the lens is 2cm from the eye

Answers

Answer:

The value is $p = - 2.63 \ Diopters$

Explanation:

From the question we are told that

The value of the far point is $a = 40 \ cm = 0.4 \ m$

The distance of the lens to the eye is $b = 2 \ cm = 0.02$

Generally

$1 Diopter = > 1 m^(-1)$

Generally the power spectacle lens needed is mathematically represented as

$p = (1)/(d_o ) + (1)/(d_i)$

Here $d_o$ is the object distance which for a near sighted person is $d_o = \infty$

And $d_i$ is the image distance which is evaluated as

$d_i = b - a$

=> $d_i = 0.02 - 0.4$

=> $d_i = -0.38 \ m$

$p = (1)/(\infty ) + (1)/(-0.38)$

=> $p = 0 - 2.63$

=> $p = - 2.63 \ Diopters$

In an experiment, one of the forces exerted on a proton is F⃗ =−αx2i^, where α=12N/m2. What is the potential-energy function for F⃗ ? Let U=0 when x=0. Express your answer in terms of α and x.

Answers

Answer

$\Delta U= \alpha (x^3)/(3) \n$

Explanation:

given

$F = -\alpha x^2 i$

where $\alpha = 12 N/m^2$

now we know

$\int\limits^W_0 {} \, dW = \int\limits^a_b {F.} \, dxi$ ..................(i)

where dx is infinitesimal distance

$W = \int\limits^a_b {-\alpha x^2} \, dx \n$

for x = a and b = 0

after integration we get

$W = -\alpha (x^3)/(3)$

we know work done by conservative force will be equals to negative of potential energy

$W = -\Delta U$

so we get

$-\Delta U= -\alpha (x^3)/(3) \n\n\Delta U= \alpha (x^3)/(3) \n$

Two ice skaters, Lilly and John, face each other while at rest, and then push against each other's hands. The mass of John is twice that of Lilly. How do their speeds compare after they push off? Lilly's speed is one-fourth of John's speed. Lilly's speed is the same as John's speed. Lilly's speed is two times John's speed. Lilly's speed is four times John's speed. Lilly's speed is one-half of John's speed.

Answers

Answer:

Lilly's speed is two times John's speed.

Explanation:

m = Mass

a = Acceleration

t = Time taken

u = Initial velocity

v = Final velocity

The force they apply on each other will be equal

$F=ma\n\Rightarrow a_l=(F)/(m_l)$

$F=ma\n\Rightarrow a_j=(F)/(2m_l)\n\Rightarrow a_j=(1)/(2)a_l$

$v=u+at\n\Rightarrow v_l=0+(F)/(m_l)* t\n\Rightarrow v_l=a_lt$

$v=u+at\n\Rightarrow v_l=0+(F)/(2m_l)* t\n\Rightarrow v_j=(1)/(2)a_lt\n\Rightarrow v_j=(1)/(2)v_l\n\Rightarrow v_l=2v_j$

Hence, Lilly's speed is two times John's speed.

Answer:

Lilly's speed is 2 times Johns speed

Explanation:

A block slides from rest with negligible friction down the track above, descending a vertical height of 5.0 m to point P at the bottom. It then slides on the horizontal surface. The coefficient of friction between the block and the horizontal surface is 0.20. How far does the block slide on the horizontal surface before it comes to rest?

Answers

The block slide on the horizontal surface is "24.99 m" far.

According to the question,

Vertical height = 5.0 m
Coefficient of friction = 0.20

Let,

The time taken be "t".

Now,

→ $s = ut+ (1)/(2) at^2$

By substituting the values, we get

$5 = (1)/(2)* 9.8* t^2$

$t = 1.01 \ sec$

The final velocity will be:

→ $v_1 = gt$

$= 9.8* 1.01$

$= 9.899 \ m/s$

Now,

→ $t = (u)/(a)$

$= (9.899)/(0.2* 9.8)$

$= 5.05 \ seconds$

hence,

The distance will be:

→ $s = ut+0.5* at^2$

$= 9.899(5.05)-0.5* (0.2* 9.8* 5.05^2)$

$= 24.99 \ m$

Thus the above approach is right.

Learn more about friction here:

brainly.com/question/18851133

Answer:

The block slides on the horizontal surface 25 m before coming to rest.

Explanation:

Hi there!

For this problem, we have to use the energy-conservation theorem. Initially, the block has only gravitational potential energy (PE) that can be calculated as follows:

PE = m · g · h

Where:

m = mass of the block.

g = acceleration due to gravity.

h = height at which the block is located.

As the block starts to slide down the track, its height diminishes as well as its potential energy. Due to the conservation of energy, energy can´t disappear, so the loss of potential energy is compensated by an increase of kinetic energy (KE). In other words, as the block slides, the potential energy is converted into kinetic energy. The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

m = mass of the block.

v = speed of the block.

Then, at the bottom of the ramp, the kinetic energy of the block will be equal to the potential energy that the block had at the top of the ramp.

Initial PE = KE at the bottom

When the block starts sliding horizontally, friction force does work to stop the block. According to the energy-work theorem, the change in the kinetic energy of an object is equal to the net work done on that object. In other words, the amount of work needed to stop the block is equal to its kinetic energy. Then, the work done by friction will be equal to the kinetic energy of the block at the bottom, that is equal to the potential energy of the block at the top of the track:

initial PE = KE at the bottom = work done by friction

The work done by friction is calculated as follows:

W = Fr · Δx

Where:

W = work

Fr = friction force.

Δx = traveled distance.

And the friction force is calculated as follows:

Fr = μ · N

Where:

μ = coefficient of friction.

N = normal force.

Since the block is not accelerated in the vertical direction, in this case, the normal force is equal to the weight (w) of the block:

Sum of vertical forces = ∑Fy = N - w = 0 ⇒N = w

And the weight is calculated as follows:

w = m · g

Where m is the mass of the block and g the acceleration due to gravity.

Then, the work done by friction can be expressed as follows:

W = μ · m · g · Δx

Using the equation:

intial PE = work done by friction

m · g · h = μ · m · g · Δx

Solving for Δx

h/μ = Δx

5.0 m / 0.20 = Δx

Δx = 25 m

The block slides on the horizontal surface 25 m before coming to rest.

Twopucksofequalmasscollideonafrictionlesssurface,asillustratedinthefigure.Immediatelyafterthe collision, the speed of the black puck is 1.5 m/s. What is the speed of the white puck immediately after the collision?

Answers

Answer:

The speed of the white puck immediately after the collision is 2.6 m/s.

Explanation:

Given that,

Two pucks are equal masses.

Speed of black puck = 1.5 m/s

According to given figure,

We need to calculate the speed of the white puck immediately after the collision

Using law of conservation of momentum

$mv=m_(1)v_(1)\cos\theta+m_(2)v_(2)\cos\theta$

Put the value into the formula according to figure

$m*3=m* v_(1)*\cos30+m*1.5*\cos60$

$3m=0.866m v_(1)+0.75m$

$v_(1)=(3-0.75)/(0.866)$

$v_(1)=2.6\ m/s$

Hence, The speed of the white puck immediately after the collision is 2.6 m/s.

A train travels due south at 20 m/s. It reverses its direction and travels due north at 20 m/s. What is the change in velocity of the train? 50 m/s, due south e50 m/s, due north 120 m/s, due south zero ms 40 m/s, due north

Answers

Answer:

40 m/s due north

Explanation:

Consider that the south direction a negative Y axis and north direction as + Y axis

v1 = 20 m/s South = 20 (-j) m/s

v2 = 20 m/s North = 20 j m/s

Change in velocity = v2 - v1 = 20 j - 20 (-j) = 40 j m/s

So, change in velocity is 40 m/s due north.

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