Physics, 28.06.2020 02:01 QueenThao8933

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?

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Answer from: csciahetano

(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:

$sin\theta=\frac{0.74m}{2.8m}=0.264\\\\\theta=sin^{-1}(0.264)=15.32\°$

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

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

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.

Answer from: Quest

electromagnetic shielding is the practice of reducing the electromagnetic field in a space by blocking the field with barriers made of conductive or magnetic materials. shielding is typically applied to enclosures to isolate electrical devices from their surroundings, and to cables to isolate wires from the environment through which the cable runs. electromagnetic shielding that blocks radio frequency electromagnetic radiation is also known as rf shielding.

the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia

Answer from: Quest

d. platinum would have the greatest temperature increase because it has the lowest specific heat.

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A 39 kg block of ice slides down a frictionless incline 2.8 m along the diagonal and 0.74 m high. A...

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