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3 years ago

Consider a force of 57.3 N, pulling 3 blocks of

Physics

1 answer:

andrezito [222]3 years ago

3 0

Block 1 (the rightmost block) has

• net horizontal force

∑ F = F - T₁ - f₁ = m₁a

• net vertical force

∑ F = N₁ - m₁g = 0

where F = 57.3 N, T₁ is the tension in the string connecting blocks 1 and 2, f₁ is the magnitude of kinetic friction felt by block 1, m₁ = 0.8 kg is its mass, a is the acceleration you want to find, and N₁ is the magnitude of the normal force exerted by the surface.

Block 2 (middle) has much the same information:

• net horiz. force

∑ F = T₁ - T₂ - f₂ = m₂a

• net vert. force

∑ F = N₂ - m₂g = 0

with similarly defined symbols.

The same goes for block 3 (leftmost):

• net horiz. force

∑ F = T₂ - f₃ = m₃a

• net vert. force

∑ F = N₃ - m₃g = 0

We have m₁ = m₂ = m₃ = 0.8 kg, so I'll just replace each with m. It follows that each normal force has the same magnitude, N₁ = N₂ = N₃ = mg. And as a consequence of that, each frictional force has the same magnitude, f₁ = f₂ = f₃ = 0.4mg.

In short, the relevant equations are

[1] … 57.3 N - T₁ - 0.4mg = ma

[2] …T₁ - T₂ - 0.4mg = ma

[3] … T₂ - 0.4mg = ma

Adding [1], [2] and [3] together eliminates the tension forces, and we get

57.3 N - 1.2mg = 3ma

Solve for a :

57.3 N - 1.2 (0.8 kg) (9.8 m/s²) = 3 (0.8 kg) a

57.3 N - 9.408 N = (2.4 kg) a

a = (47.892 N) / (2.4 kg)

a ≈ 20.0 m/s²

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The world speed record on water was set on October 8, 1978 by Ken Warby. If Ken drove his motorboat a distance of 1500 m in 8.10

maxonik [38]

Answer:

317.52 mi/hr

Explanation:

First convert Meters into miles as the answer is required in miles/ h

1000m = 0.62 mi

Now, convert second into hours

7.45s = 0.0001 hr

The speed of the boat would be

v = 0.62/0.0001

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3 years ago

A 95kg fullback (football player for those not into sports) moving south with a speed of 5.0 m/s has a perfectly inelastic colli

Lunna [17]

Answer:

a. $v=3.11mls$ , $29.4^{0}$

b. $K.E =-697.8J$

Explanation:

To calculate the values in the question, a deep understanding of perfect inelastic collision is important.

When two bodies undergo inelastic collision, two important parameters must be well understood i.e

Momentum: the momentum is always conserved in perfectly inelastic collision. i.e the total momentum after collision is the sum of the individual momentum before collision

Kinetic energy: Kinetic energy is not conserved due to dissipative force.

a.To calculate the velocity, we first find the total momentum before collision

Momentum of player 1 $p_{1} =mv=95kg*5m/s\\p_{1} =475kgm/s\\$

Momentum of player 2 $p_{2} =mv=90kg*3m/s\\p_{1} =270kgm/s\\$

Hence the total momentum $p_{12}=p_{1}+p_{2}\\$

Note, since the direction of movement before collision is due south and due north respectively we have to represent the velocity using the rectangular coordinate

Hence $p_{12}=(m_{1}+m_{2})v=p_{1}i+p_{2}j\\$

$(95+90)v=475i+270j\\$

$v=2.57i+1.45j\\$

solving for the resultant velocity, we have

$v=\sqrt{2.75^{2} +1.45^{2}}\\ v=3.11mls$

To calculate the direction of movement, we have

$\alpha =tan^{-1}=\frac{v_{j} }{v_{i}}\\ \alpha =tan^{-1}=\frac{1.45}{2.57}\\\alpha =29.4^{0}$

b. to calculate the decrease in total kinetic energy, before collision, the total kinetic was

$K.E_{initial} =\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}.\\K.E_{initial} =((1/2)*95*5^{2})+((1/2)*90*3^{2})\\K.E_{initial} =1187.5+405\\K.E_{initial} =1592.5J\\$

And the final kinetic energy after collision is

$K.E_{final} =\frac{1}{2}(m_{1}+m_{2} )v^{2}\\ K.E_{final} =\frac{1}{2}(95+90)* 3.11^{2}\\ K.E_{final} =894.7J$

The decrease in Kinetic energy is

$K.E =K.E_{final}- K.E_{initial}=894.7-1592.5$

$K.E =-697.8J$

The negative sign indicate a decrease in Kinetic energy

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3 years ago

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3 years ago

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The weights in atwoods machine, starting at rest, attain a velocity of 2ft/sec in one sec. Find the ratio of the masses

Orlov [11]

Refer to the figure shown below.
Let m₁ and m₂ e the two masses.
Let a = the acceleration.
Let T = tension over the frictionless pulley.

Write the equations of motion.
m₂g - T = m₂a (1)
T - m₁g = m₁a (2)

Add equations (1) and (2).
m₂g - T + T - m₁g = (m₁ + m₂)a
(m₂ - m₁)g = (m₁ + m₂)a

Divide through by m₁.
(m₂/m₁ - 1)g = (1 + m₂/m₁)a

Define r = m₂/m₁ as the ratio of the two masses. Then
(r - 1)g = (1 +r)a
r(g-a) = a + g
r = (g - a)/(g + a)

With = 2 ft/s from rest, the acceleration is
a = 2/32.2 = 0.062 ft/s²
Therefore
r = (32.2 - 0.062)/(32.2 + 0.062) = 0.9962

Answer:
The ratio of masses is 0.9962 (heavier mass divided by the lighter mass).