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Lady_Fox [76]
4 years ago
14

A truck pulls a large box to the left with two large chains. The truck is unable to move the box because the large box is in equ

ilibrium. One of the chains has tension in the amount of 1000 Newtons. If the force of friction between the box and the ground is 2500 Newtons, explain and solve for what the tension is in the other chain. Use any relevant rules, laws, and show your work using any math reasoning.

Physics
1 answer:
iren [92.7K]4 years ago
4 0
Refer to the diagram shown below.

m = the mass of the crate.
W = mg, the weight of the crate (g = 9.8 m/s²).
N = the normal reaction from the ground.

R =  2500 N, the frictional force between the crate and the ground.
R = μN = μmg
      where
     μ =  the static coefficient of friction

The force in the first chain is 1000 N.
Let T =  the force in the second chain.
In order to move the crate, the applied force should exceed the resisting frictional force. That is,
T + 1000 ≥ 2500
T ≥ 1500 N
Because the crate does not move, T = 1500 N.

Answer: T = 1500 N

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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
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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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Answer:(4).

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What are three properties of electromagnetic waves?
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The three properties of electromagnetic waves are; they travel at the speed of light, they include ultraviolet waves, and they can transfer energy through empty space.

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Subject: Physics

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Sub-topic: Properties of an electromagnetic waves

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4 years ago
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A child is holding a wagon from rolling straight back down in a driveway that inclined at 20 degree horizontal. if the wagon wei
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Answer:

F = 51.3°

Explanation:

The component of weight parallel to the inclined plane must be responsible for the rolling back motion of the car. Hence, the force required to be applied by the child must also be equal to that component of weight:

F = Parallel\ Component\ of\ Weight\ of\ Wagon= WSin\theta\\

where,

W = Weight of Wagon = 150 N

θ = Angle of Inclinition = 20°

Therefore,

F = (150\ N)Sin\ 20^o

<u>F = 51.3°</u>

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3 years ago
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