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liq [111]
2 years ago
8

A large box of mass M is pulled across a horizontal, frictionless surface by a horizontal rope with tension T. A small box of ma

ss m sits on top of the large box. The coefficients of static and kinetic friction between the two boxes are μs and μk, respectively.
1- Find an expression for the maximum tension T max for which the small box rides on top of the large box without slipping.Express your answer in terms of the variablesM ,m , ms, and appropriate constants.2-A horizontal rope pulls a 10 kg wood sled across frictionless snow. A 6.0 kg wood box rides on the sled. What is the largest tension force for which the box doesn't slip? Assume that Mk= 0.50.
Physics
1 answer:
Nesterboy [21]2 years ago
5 0

Answer:

T = g μ_s ( M+m )

78.4 N

Explanation:

When both of them move with the same acceleration , small box will not slip over the bigger one. When we apply force on the lower box, it starts moving with respect to lower box. So a frictional force arises on the lower box which helps it too to go ahead . The maximum value that this force can attain is mg μ_s . As a reaction of this force, another force acts on the lower box in opposite direction .

Net force on the lower box

= T - mg μ_s = M a    ( a is the acceleration created by net force in M )

Considering force on the upper box

mg μ_s = ma

a = g μ_s

Put this value of a in the equation above

T - m gμ_s = M g μ_s

T = mg μ_s + M g μ_s

=  g μ_s ( M+m )

2 )

Largest tension required

T = 9.8 x  .50 x ( 10+6 )

= 78.4 N

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A cliff diver running 3.60 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 2.00 s later
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Explanation:

It is given that,

The horizontal speed of a cliff diver, v_x=3.6\ m/s

It reaches the water below 2.00 s later, t = 2 s

Let d_x is the distance where the diver hit the water. It can be calculated as follows :

d_x=v_x\times t\\\\=3.6\times 2\\\\=7.2\ m

Let d_y is the height of the cliff. It can be calculated using second equation of motion as follows :

d_y=u_yt+\dfrac{1}{2}gt^2\\\\d_y=\dfrac{1}{2}\times 9.8\times 2^2\\\\=19.6\ m

So, the cliff is 19.6 m high and it will hit the water at a distance of 19.6 m.

8 0
1 year ago
A block rests on a flat plate that executes vertical simple harmonic motion with a period of 0.74 s. What is the maximum amplitu
Mumz [18]

Answer:

maximum amplitude  = 0.13 m

Explanation:

Given that

Time period T= 0.74 s

acceleration of gravity g= 10 m/s²

We know that time period of simple harmonic motion given as

T=\dfrac{2\pi}{\omega}

0.74=\dfrac{2\pi}{\omega}

ω = 8.48 rad/s

ω=angular frequency

Lets take amplitude = A

The maximum acceleration given as

a= ω² A

The maximum acceleration should be equal to g ,then block does not separate

a= ω² A

10= 8.48² A

A=0.13 m

maximum amplitude  = 0.13 m

8 0
2 years ago
Question 2 (1 point)
love history [14]

Answer:

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Explanation:

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6 0
2 years ago
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13) Can two nonzero perpendicular vectors be added together so their sum is zero? Explain.
Neporo4naja [7]

In order for two vectors to add to zero, they must have the same magnitude and point in opposite directions.

Two perpendicular vectors, by definition, make a right angle with each other whereas two vectors pointing in opposite directions form a straight line.

Because of this, two perpendicular vectors with nonzero magnitudes will never add to zero.

3 0
2 years ago
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A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block
melomori [17]

This question is incomplete, the complete question is;

A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block slides down the ramp over a distance d until it reaches the bottom of the ramp.

How much of its original total energy (in J) survives as KE when it reaches the ground? m = 9.9 kg h = 4.9 m d = 5 m μ = 0.3 θ = 36.87°

Answer:

the amount of its original total energy (in J) that survives as KE when it reaches the ground will is 358.975 J

Explanation:

Given that;

m = 9.9 kg

h = 4.9 m

d = 5 m

μ = 0.3

θ = 36.87°

Now from conservation of energy, the energy is;

Et = mgh

we substitute

Et = 9.9 × 9.8 × 4.9

= 475.398 J

Also the loss of energy i

E_loss = (umg cosθ) d

we substitute

E_loss  = 0.3 × 9.9 × 9.8 × cos36.87°  × 5

= 116.423 J

so the amount of its original total energy (in J) that survives as KE when it reaches the ground will be

E = Et - E_loss

E = 475.398 J - 116.423 J

E = 358.975 J

5 0
2 years ago
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