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3 years ago

7

In a new scenario, the block only makes it (exactly) half-way through the rough spot. How far was the spring compressed from its

unstretched length

1 answer:

Mila [183]3 years ago

4 0

Answer: hello below is the missing part of your question

A mass m = 10 kg rests on a frictionless table and accelerated by a spring with spring constant k = 5029 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is μk = 0.49. The mass leaves the spring at a speed v = 3.4 m/s.

answer

x = 0.0962 m

Explanation:

<em>First step :</em>

Determine the length of the rough patch/spot

F = Uₓ (mg)

and w = F.d = Uₓ (mg) * d

hence;

d( length of rough patch) = w / Uₓ (mg) = 46.55 / (0.49 * 10 * 9.8) = 0.9694 m

<em>next : </em>

work done on unstretched spring length

Given that block travels halfway i.e. d = 0.9694 / 2 = 0.4847 m

w' = Uₓ (mg) * d

= 0.49 * 10 * 9.81 * 0.4847 = 23.27 J

also given that the Elastic energy of spring = work done ( w')

1/2 * kx^2 = 23.27 J

x = $\sqrt{\frac{2*23.27}{5029} }$ = 0.0962 m

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Answer:

1.56 J

Explanation:

The potential energy only depends on the vertical height from the ground level.

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4 years ago

A plane lands on a runway with a speed of 115 m/s, moving east, and it slows to a stop in 13.0 s. What is the magnitude (in m/s2

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Answer:

<em> The planes average acceleration in magnitude and direction = 8.846 m/s² moving east</em>

Explanation:

Acceleration: This can be defined as the rate of change of velocity. The S.I Unit of acceleration is m/s². Acceleration is a vector quantity because it can be represented both in magnitude and in direction.

Acceleration can be represented mathematically as

a = v/t.................................... Equation 1

Where a = acceleration, v = velocity, t= time.

<em>Given: v = 115 m/s, t = 13.0 s</em>

<em>Substituting these values into equation 1</em>

<em>a = 115/13</em>

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3 years ago

Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2

Bess [88]

Answer:

The acceleration of $M_2$ is $a = 0.7156 m/s^2$

Explanation:

From the question we are told that

The mass of first block is $M_1 = 2.25 \ kg$

The angle of inclination of first block is $\theta _1 = 43.5^o$

The coefficient of kinetic friction of the first block is $\mu_1 = 0.205$

The mass of the second block is $M_2 = 5.45 \ kg$

The angle of inclination of the second block is $\theta _2 = 32.5^o$

The coefficient of kinetic friction of the second block is $\mu _2 = 0.105$

The acceleration of $M_1 \ and\ M_2$ are same

The force acting on the mass $M_1$ is mathematically represented as

$F_1 = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

=> $M_1 a = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

Where T is the tension on the rope

The force acting on the mass $M_2$ is mathematically represented as

$F_2 = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

$M_2 a = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

At equilibrium

$F_1 = F_2$

So

$T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

making a the subject of the formula

$a = \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}$

substituting values $a = \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}$

=> $a = 0.7156 m/s^2$

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