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bulgar [2K]
3 years ago
10

A box of mass 3.1kg slides down a rough vertical wall. The gravitational force on the box is 30N . When the box reaches a speed

of 2.5m/s , you start pushing on one edge of the box at a 45? angle (use degrees in your calculations throughout this problem) with a constant force of magnitude Fp = 23N , as shown in figure. (Figure 1) There is now a frictional force between the box and the wall of magnitude 13N . How fast is the box sliding 2.0s after you started pushing on it?
Using our simplified model, in which we know that the forces are constant (but we don't know what their magnitudes are), which of the following motion diagrams would be a reasonable representation of the motion of the box?
Physics
2 answers:
hoa [83]3 years ago
6 0

Answer:

The speed of the box after 2 s is 3 m/s.

Explanation:

The net force on the box in vertical direction:

{F_{net}} = {F_g} - f - {F_p}\sin {45^{\rm{o}}}F

Here, {F_g} is the gravitational force, f is the force of friction and {F_p} is the pushing force.

According to Newton’s second law, the net force is:

{F_{net}} = ma

Here, a is the acceleration in vertical direction

\begin{array}{c}\\ \Rightarrow ma = {F_g} - f - {F_p}\sin {45^{\rm{o}}}\\\\a = \frac{{{F_g} - f - {F_p}\sin {{45}^{\rm{o}}}}}{m}\\\end{array}

Substitute 30N for Fg, 13N for f and 23N for Fp

 \begin{array}{c}\\a = \frac{{\left( {30{\rm{ N}}} \right) - \left( {13{\rm{ N}}} \right) - \left( {23{\rm{ N}}} \right)\sin {{45}^{\rm{o}}}}}{{3.1{\rm{ kg}}}}\\\\ = \frac{{0.7365{\rm{ N}}}}{{3.1{\rm{ kg}}}}\\\\ = 0.24\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\\\end{array}

Newton’s equation of motion:

v - u = at

Substitute 2.5m/s for u , 0.24m/s2 for a and 2s for t .

\begin{array}{c}\\v - 2.5{\rm{ m/s}} = \left( {0.24{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {2{\rm{ s}}} \right)\\\\v = 0.48{\rm{ m/s}} + 2.5{\rm{ m/s}}\\\\ = 2.98{\rm{ m/s}}\\\\ \approx {\bf{3 m}}{\rm{/}}{\bf{s}}\\\end{array}

The speed of the box after 2 s is 3 m/s.

Tems11 [23]3 years ago
5 0

Answer:

The box will be moving at 0.45m/s. The solution to this problem requires the knowledge and application of newtons second law of motion and the knowledge of linear motion. The vertical component of the force Fp acts vertically upwards against the directio of motion. This causes a constant upward force of 23sin45° to act on the box. Fhe frictional force of 13N also acts vertically upwards and so two forces act upwards against rhe force of gravity resulting un a net force of 0.7N acting kn the box. This corresponds to an acceleration of 0.225m/s². So in w.0s after i start to push v = 0.45m/s.

Explanation:

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Hardness and density are physical properties.<br> a. True<br> b. False
antiseptic1488 [7]
a . true hardness and density are physical properties
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If both of these balls are traveling down a bowling lane at the SAME speed, which one would would have more kinetic energy?
Mariana [72]
<h2>Answer:</h2>

<u>Ball A has more kinetic energy</u>

<h2>Explanation:</h2>

As we know that Kinetic energy is given by

K. E = 1/2 mv²

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As the mass of ball A is greater than ball B so we can say that the kinetic energy of ball is more than ball B


3 0
3 years ago
Determine the direction of the force that will act on the charge in each of the following situations. A negative charge moving t
wlad13 [49]

Answer:

a) DOWN direction,  b)  directed INTO THE SCREEN, c)    F = 0

Explanation:

The direction of the force is

for electric force

           F = q E

where we assume a positive test charge, for which the force has the direction of the electric field.

For a magnetic field

in this case the direction of the force is given by the right hand rule.

For a positive test charge, the thumb points in the direction of velocity, the other fingers extended in the direction of the magnetic field, and the palm gives the direction of force for a positive charge.

           F = q v x B

Let us apply these considerations to our case.

a) negative charge moving to the left

in a magnetic field points away from the screen

In this case the thumb goes to the left, the fingers extended outwards and the palm points upwards, but since the charge is negative the force has a DOWN direction.

b) negative charge moves to the left

in electric field it points off the screen.

The outside is in the direction of the electric field and since the charge is negative, the force is directed INTO THE SCREEN

c) positive charge moves down

in magnetic field points up

in this case the velocity and the field have the same direction so the vector product of them is zero

       F = q v  B sin 0

       F = 0

6 0
3 years ago
An 3.7 lb hammer head, traveling at 5.8 ft/s strikes a nail and is brought to a stop in 0.00068 s. The acceleration of gravity i
CaHeK987 [17]

Answer:

31677.2 lb

Explanation:

mass of hammer (m) = 3.7 lb

initial velocity (u) = 5.8 ft/s

final velocity (v) = 0

time (t) = 0.00068 s

acceleration due to gravity (g) 32 ft/s^{2}

force = m x ( a + g )

where

  • m is the mass = 3.7 lb
  • g is the acceleration due to gravity = 32 ft/s^{2}
  • a is the acceleration of the hammer

       from v = u + at

       a = (v-u)/ t

       a = (0-5.8)/0.00068 = -8529.4 ( the negative sign showa the its decelerating)

we can substitute all required values into force= m x (a+g)

force = 3.7 x (8529.4 + 32) = 31677.2 lb

       

4 0
3 years ago
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