PHYSICS COLLEGE

A 39 kg block of ice slides down a frictionless incline 2.8 m along the diagonal and 0.74 m high. A worker pushes up against the ice, parallel to the incline, so that the block slides down at constant speed. (a) Find the magnitude of the worker's force. How much work is done on the block by (b) the worker's force, (c) the gravitational force on the block, (d) the normal force on the block from the surface of the incline, and (e) the net force on the block?

Answers

Answer 1

Answer:

Answer:

(a) Fw = 101.01 N

(b) W = 282.82 J

(d) N = 368.61 N

(e) Net force = 0 N

Explanation:

(a) In order to calculate the magnitude of the worker's force, you take into account that if the ice block slides down with a constant speed, the sum of forces, gravitational force and work's force, must be equal to zero, as follow:

$F_g-F_w=0$ (1)

Fg: gravitational force over the object

Fw: worker's force

However, in an incline you have that the gravitational force on the object, due to its weight, is given by:

$F_g=Wsin\theta=Mg sin\theta$ (2)

M: mass of the ice block = 39 kg

g: gravitational constant = 9.8m/s^2

θ: angle of the incline

You calculate the angle by using the information about the distance of the incline and its height, as follow:

$sin\theta=(0.74m)/(2.8m)=0.264\n\n\theta=sin^(-1)(0.264)=15.32\°$

Finally, you solve the equation (1) for Fw and replace the values of all parameters:

$F_w=F_g=Mgsin\theta\n\nF_w=(39kg)(9.8m/s^2)sin(15.32\°)=101.01N$

The worker's force is 101.01N

(b) The work done by the worker is given by:

$W=F_wd=(101.01N)(2.8m)=282.82J$

(c) The gravitational force on the block is, without taking into account the rotated system for the incline, only the weight of the ice block:

$F_g=Mg=(39kg)(9.8m/s^2)=382.2N$

The gravitational force is 382.2N

(d) The normal force is:

$N=Mgcos\theta=(39kg)(9.8m/s^2)cos(15.32\°)=368.61N$

(e) The speed of the block when it slides down the incle is constant, then, by the Newton second law you can conclude that the net force is zero.

Related Questions

Which statements explain why theories change over time? Check all that apply. Scientists change the definition of theory to have their ideas accepted. New information and technology may be developed that influence the theory. Theories change with each new generation of scientists. New experimental methods provide new information. Theories may or may not be supported by new information.
We know that there is a relationship between work and mechanical energy change. Whenever work is done upon an object by an external force (or non-conservative force), there will be a change in the total mechanical energy of the object. If only internal forces are doing work then there is no change in the total amount of mechanical energy. The total mechanical energy is said to be conserved. Think of a real-life situation where we make use of this conservation of mechanical energy (where we can neglect external forces for the most part). Describe your example and speak to both the kinetic and potential energy of the motion.
An AC voltage source has an output of ∆V = 160.0 sin(495t) Volts. Calculate the RMS voltage. Tries 0/20 What is the frequency of the source? Tries 0/20 Calculate the voltage at time t = 1/106 s. Tries 0/20 Calculate the maximum current in the circuit when the generator is connected to an R = 53.8 Ω resistor.
Convert 7 (gcm^2)/(min^2) into a value in standard S.I. units. Be sure to use scientific notation if necessary. You do not need to answer units.
A 50-gram ball is released from rest 80 m above the surface of the Earth. During the fall to the Earth, the total thermal energy of the ball and the air in the system increases by 15 J. Just before it hits the surface its speed is

determine exactly where to place a cart on the track so that it rolls down the track, flies through the air, and lands precisely at 1) the green line, 2) the red line, and 3) the blue line, on the first try.

Answers

Answer: i think you should place it on the red line

Explanation:

hope this helps

and need brainliest

The position of a particle changes linearly with time, i.e. as one power of t, as given by the following: h(t) = ( 4.1 t + 5.5 ) meters. Find the speed of the particle, in meters per second.

Answers

Answer:

v = 4.1 m / s

Explanation:

Velocity is defined by the relation

v = $(dx)/(dt)$

we perform the derivative

v = 4.1 m / s

Another way to find this magnitude is to see that the velocity on the slope of a graph of h vs t

v = $(\Delta x)/(\Delta t)$

Δx = v Δdt + x₀

h= 4.1 t + 5.5

v = 4.1 m / s

x₀ = 5.5 m

The Speed of a Particle is 4.1 meters per second.

The position of a particle can be represented by a linear equation of the form h(t) = (at + b) where a and b are constants.

In this case, the equation is h(t) = (4.1t + 5.5).

To find the speed of the particle, we can take the derivative of the position equation with respect to time.

The derivative of h(t) is the rate of change of position with respect to time, which represents the velocity of the particle.

In this case, the derivative is 4.1 meters per second.

Therefore, the speed of the particle is 4.1 meters per second.

Learn more about Speed of a Particle here:

brainly.com/question/32295338

#SPJ3

Which statementsabout a neutral atom are correct? Check all that apply.1. A neutral atom is composed of bothpositively and negatively charged particles.
2. Positively charged protons are located in the tiny, massivenucleus.
3. The positively chargedparticles in the nucleus are positrons.
4. The negatively chargedelectrons are spread out in a "cloud" around thenucleus.
5. The electrons areattracted to the positively charged nucleus.
6. The radius of the electroncloud is twice as large as the radius of the nucleus.

Answers

Answer:

1, 2, 4, 5 are correct

Explanation:

1) This is true because In a neutral atom, the number of positive charges (protons) is equal to the number of negative charges (electrons).

2) This is true because the mass of the atom which is made up of the protons and neutrons, is located in the tiny nucleus.

3) This is not true because the positively charged particles in the nucleus are called protons.

4) This is true because electrons move around the nucleus in diffuse areas known as orbitals.

5) This is true because opposite charges attract each other. And electron is a negative charge.

6) This is not true because the radius of the electron cloud is normally 10,000 times larger than the radius of the nucleus.

The uniform crate has a mass of 50 kg and rests on the cart having an inclined surface. Determine the smallest acceleration that will cause the crate either to tip or slip relative to the cart. What is the magnitude of this acceleration

Answers

Answer:

The answer is below

Explanation:

Let g = acceleration due to gravity = 9.81 m/s², x = half of the width of the crate, half of the height of the crate = 0.5 m, a = acceleration of crate, N = force raising the crate

The sum of moment is given as:

$50asin(15)x+50acos(15)0.5=-50(9.81)sin(15)0.5+50(9.81)cos(15)x\ \ \ (1)$

Sum of vertical forces is zero, hence:

$N-50(9.81)cos(15)+50acos(15)=0\ \ \ (2)$

Sum of horizontal force is zero, hence:

$50(9.81)sin(15)-\mu N+50acos(15)=0\n\n50(9.81)sin(15)-0.5 N+50acos(15)=0\ \ \ (3)$

Solving equation 1, 2 and 3 simultaneously gives :

N = 447.8 N, a = 2.01 m/s², x = 0.25 m

x is supposed to be 0.3 m (0.6/2)

The crate would slip because x <0.3 m

Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is a distance of 4.84 mm from the center of the central bright fringeFor what wavelength of light will thefirst-order dark fringe be observed at this same point on thescreen?
Express your answer in micrometers(not in nanometers).

Answers

Answer:

1.196 μm

Explanation:

D = Screen distance = 3 m

$\lambda$ = Wavelength = 598 m

y = Distance of first-order bright fringe from the center of the central bright fringe = 4.84 mm

d = Slit distance

$tan\theta=(y)/(D)\n\Rightarrow \theta=tan^(-1){(y)/(D)}\n\Rightarrow \theta=tan^(-1){(4.84* 10^(-3))/(3)}\n\Rightarrow \theta=0.09243\ ^(\circ)$

$sin\theta=(\lambda)/(d)\n\Rightarrow d=(\lambda)/(sin\theta)\n\Rightarrow d=(598* 10^(-9))/(sin0.09243)\n\Rightarrow d=0.00037066\ m$

For first dark fringe

$dsin\theta=(\lambda')/(2)\n\Rightarrow \lambda'=2dsin\theta\n\Rightarrow \lambda'=2* 0.00037066* sin0.09243\n\Rightarrow \lambda'=1.196* 10^(-6)\n\Rightarrow \lambda'=1.196\ \mu m$

Wavelength of first-order dark fringe observed at this same point on the screen is 1.196 μm

Final answer:

The wavelength of light that will produce the first-order dark fringe at the same point on the screen is the same as the original wavelength of the light, which is 598 nm (0.598 μm).

Explanation:

To find the wavelength of light that will produce the first-order dark fringe at the same point on the screen, we can use the equation dsinθ = nλ, where d is the separation between the slits, θ is the angle of the fringe, n is the order of the fringe, and λ is the wavelength of the light.

In this case, the first-order bright fringe is located at a distance of 4.84 mm from the center of the central bright fringe. Since this is a first-order fringe, n = 1.

Plugging in the values, we have (0.120 mm)(sinθ) = (1)(λ). Rearranging the equation gives sinθ = λ/0.120 mm.

Since the first-order dark fringe is located at the same point as the first-order bright fringe, the angle of the first-order dark fringe can be calculated by taking the sine inverse of λ/0.120 mm.

Finally, to find the wavelength of light that will produce the first-order dark fringe at this point, we can rearrange the equation to solve for λ: λ = (0.120 mm)(sinθ).

Now, substitute the known values into the equation to calculate the wavelength of light:

λ = (0.120 mm)(sinθ) = (0.120 mm)(sin sin^-1(λ/0.120 mm)) = λ.

The wavelength of light that will produce the first-order dark fringe at this point on the screen is the same as the original wavelength of light, which is 598 nm. Converting this value to micrometers, we get 0.598 μm.

Learn more about interference pattern here:

brainly.com/question/33930819

#SPJ3

A parallel-plate vacuum capacitor has 7.72 J of energy stored in it. The separation between the plates is 3.30 mm. If the separation is decreased to 1.45 mm, For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Stored energy. Part A what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed

Answers

Answer

3.340J

Explanation;

Using the relation. Energy stored in capacitor = U = 7.72 J

U =(1/2)CV^2

C =(eo)A/d

C*d=(eo)A=constant

C2d2=C1d1

C2=C1d1/d2

The separation between the plates is 3.30mm . The separation is decreased to 1.45 mm.

Initial separation between the plates =d1= 3.30mm .

Final separation = d2 = 1.45 mm

(A) if the capacitor was disconnected from the potential source before the separation of the plates was changed, charge 'q' remains same

Energy=U =(1/2)q^2/C

U2C2 = U1C1

U2 =U1C1 /C2

U2 =U1d2/d1

Final energy = Uf = initial energy *d2/d1

Final energy = Uf =7.72*1.45/3.30

(A) Final energy = Uf = 3.340J

Other Questions

A 0.26 kg rock is thrown vertically upward from the top of a cliff that is 32 m high. When it hits the ground at the base of the cliff, the rock has a speed of 29 m/s. Assuming that air resistance can be ignore and using conservation of mechanical energy, find (a) the initial speed of the rock and (b) the greatest height of the rock as measured from the base of the cliff.

Assume the following values: d1 = 0.880 m , d2 = 1.11 m , d3 = 0.560 m , d4 = 2.08 m , F1 = 510 N , F2 = 306 N , F3 = 501 N , F4 = 407 N , and MA = 1504 N⋅m . Express the Cartesian components of the resultant force and the couple moment in newtons and newton-meters to three significant figures separated by commas.

How many nanoseconds are in one hour? How do you write the following in scientific notation?2,560,000m

You wish to buy a motor that will be used to lift a 10-kg bundle of shingles from the ground to the roof of a house. The shingles are to have a 1.5-m/s2 upward acceleration at the start of the lift. The very light pulley on the motor has a radius of 0.17 m . Part A Determine the minimum torque that the motor must be able to provide. Express your answer with