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3 years ago

suppose the spring in the sample problem is replaced with a spring that stretches 36 cm from its equilibrium position

1 answer:

amm18123 years ago

3 0

The gravitational force is acting downward as the spring is held vertically. Therefore, spring is stretched downward and the force is directly proportional to the stretched distance. The product of the mass and the gravitational acceleration gives gravitational force.

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5) A 20.0 kg cart with no friction wheels sits on a table. A light string is attached to it and runs over a low friction pulley

ella [17]

Answer:

1) Please find attached, created with Microsoft Visio

2) The acceleration of the masses connected by the light string is 0.00735 m/s²

3) The tension in the cord is 0.147 N

4) The time it would take the block to go 1.2 m to the edge of the table is approximately 18.07 s

5) The velocity of the cart as soon as it gets to the edge of the table is 0.042 m/s

Explanation:

1) Please find attached, the required free body diagram, showing the tension, weight and frictional (zero friction) forces acting on the cart and the mass created with Microsoft Visio

2) The acceleration of the masses connected by the light string is given as follows;

F = Mass, m × Acceleration, a

The mass of the truck, M = 20.0 kg

The mass attached to the string, hanging rom the pulley, m = 0.0150 kg

The force, F acting on the system = The pulling force on the cart = The tension on the cable = The weight of the hanging mass = 0.0150 × 9.8 = 0.147 N

The pulling force acting on the cart, F = M × a

∴ F = 0.147 N = 20.0 kg × a

a = 0.147 N/(20.0 kg) = 0.00735 m/s²

The acceleration of the truck = a = 0.00735 m/s²

3) The tension in the cord = F = 0.147 N

4) The time, t, it would take the block to go 1.2 m to the edge of the table is given by the kinematic equation, s = u·t + 1/2·a·t²

Where;

s = The distance to the edge of the table = 1.2 m

u = The initial velocity = 0 m/s (The cart is assumed to be initially at rest)

a = The acceleration of the cart = 0.00735 m/s²

t = The time taken

Substituting the known values, gives;

s = u·t + 1/2·a·t²

1.2 = 0 × t + 1/2 ×0.00735 × t²

1.2 = 1/2 ×0.00735 × t²

t² = 1.2/(1/2 ×0.00735) ≈ 326.5306

t = √(1.2/(1/2 ×0.00735)) ≈ 18.07

The time it would take the block to go 1.2 m to the edge of the table = t ≈ 18.07 s

5) The velocity, v, of the cart as soon as it gets to the edge of the table is given by the kinematic equation, v² = u² + 2·a·s as follows;

v² = u² + 2·a·s

u = 0 m/s

v² = 0² + 2 × 0.00735 × 1.2 = 0.001764

v = √(0.001764) = 0.042

The velocity of the cart as soon as it gets to the edge of the table = v = 0.042 m/s.

7 0

3 years ago

Read 2 more answers

4. I drop a pufferfish of mass 5 kg from a height of 5.5 m onto an upright spring of total length 0.5 m and spring constant 3000

KatRina [158]

Answer:

a) 0.28 m or 28 cm is the minimum height above ground the fish reaches.

b) at the height of 0.484 m height , the pufferfish will eventually come to rest.

c) There exists two types of energy remain at the equilibrium point in the system. These are :

Gravitational potential energy = 23.72J

Spring potential energy = 0.384 J

Explanation:

Given that :

Mass of the pufferfish m =5kg

initial height of the fish h =5.5m

length of the spring l =0.5m

Spring constant K =3000N/m

Assuming no energy loss to friction, what is the minimum height above the ground that the pufferfish reaches?

Lets assume that the minimum height the fish reaches is = x meters

Now by using the conservation of energy; we realize that :

Initial total energy = final total energy

Gravitational potential energy =

Gravitational potential energy' + Spring potential energy (kinetic energy is zero in both cases)

$mgh = mgx + \frac{1}{2}K(l-x)^2$

Replacing our given values into the above equation; we have :

$(5)(9.8)(5.5) = (5)(9.5)(x) + \frac{1}{2}(3000)(0.5-x)^2$

269.5 = 47.5 x + 1500(0.5 -x )²

269.5 = 47.5 x + 1500(0.25 - x²)

269.5 = 47.5 x + 375 - 1500 x²

269.5 - 375 = 47.5 x - 1500 x²

-105.5 = 47.5 x - 1500 x²

-105.5 + 1500 x² - 47.5 x = 0

1500 x² - 47.5 x - 105.5 = 0

By using quadratic equation and taking the positive value;

x = 0.28 m or 28 cm is the minimum height above ground the fish reaches.

At the equilibrium position the weight of fish will be equal to the force applied by the spring thus

mg = kx

substituting our given values ; we have:

(5)(9.8) = 3000x

x = 61.22

x = 0.016m : so this is the compression in the spring

Now; to determine the height the pufferfish gets to before it eventually come to rest; we have

(0.5-0.016) m = 0.484m

therefore, at the height of 0.484 m height , the pufferfish will eventually come to rest.

There exists two types of energy remain at the equilibrium point in the system. These are :

Gravitational potential energy = mgh' = (5)(9.8)(0.484)

= 23.72J

and spring potential energy

$=\frac{1}{2}Kx^2\\ = \frac{1}{2}(3000)(0.016)^2\\= 0.384J$

8 0

3 years ago

A person of 70 kg standing on an un-deformable horizontal surface. She bends her knees and jumps up from rest, achieving a launc

eduard

Answer:

1190 N

Explanation:

Force: This can be defined as the product of mass and velocity. The unit of force is Newton(N).

From the question,

F = ma................. Equation 1

Where F = average force, m = mass, a = acceleration.

But,

a = (v-u)/t................ Equation 2

Where v = final velocity, u = initial velocity, t = time.

Substitute equation 2 into equation 1

F = m(v-u)/t.............. Equation 3

Given: m = 70 kg, v = 1.7 m/s, u = 0 m/s (from rest), t = 0.1 s.

Substitute into equation 3

F = 70(1.7-0)/0.1

F = 1190 N.

3 0

3 years ago

A monochromatic light passes through a narrow slit and forms a diffraction pattern on a screen behind the slit. As the slit widt

maw [93]

Answer:

shrinks with all the fringes getting narrower

Explanation:

As the light passes through the slit, the diffraction pattern shrinks, as the waves have more opening to penetrate, and the fringes becomes more narrow as a result of that, The opposite happens as the conditions are reversed.

8 0

3 years ago

An ideal refrigerator does 130. 0 j of work to remove 780. 0 j of heat from its cold compartment during each cycle. what is the

zheka24 [161]

The refrigerator's coefficient of performance is 6.

The heat extracted from the cold reservoir Q cold (i.e., inside a refrigerator) divided by the work W required to remove the heat is known as the coefficient of performance, or COP, of a refrigerator (i.e., the work done by the compressor). The required inside temperature and the outside temperature have a significant impact on the COP.

As the inside temperature of the refrigerator decreases, its coefficient of performance decreases. The coefficient of performance (COP) of refrigeration is always more than 1.

The heat produced in the cold compartment, H = 780.0 J

Work done in ideal refrigerator, W = 130.0 J

Refrigerator's coefficient of performance = H/W

= 780/130

= 6

Therefore, the refrigerator's coefficient of performance is 6.

Energy conservation requires the exhaust heat to be = 780 + 130

= 910 J

Learn more about coefficient here:

brainly.com/question/18915846

#SPJ4

5 0

2 years ago