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3 years ago

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28. The mass of a steel building frame is 5500 kg. What power is used to raise it to a helght of 5.0 m If the work is done in 12

.0 seconds?
Remember to include your data, equation, and work when solving this problem.

1 answer:

vodka [1.7K]3 years ago

8 0

Answer:

2292W

Explanation:

Use the formula for energy first:

$E=D*M$

Where E = Energy || D = Distance || M = Mass.

Solve that equation:

E = 5 * 5500 ==> 27500J

To find power, divide energy per time:

27500J / 12s =~ 2292W

Did it in a rush, any problems message me

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A 4.00 kg object is moving at 5.00 m/s NORTH. It strikes a 6.00 kg object that is moving WEST at 2.00 m/s. The objects undergo a

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We are given that an object is moving north at a speed of 5 m/s, lets call this object 1. We have another object moving west at a speed of 2 m/s, this is object 2. A diagram of the problem is the following:

To determine the loss in kinetic energy we need to determine the difference in kinetic energy before the collision and after the collision:

$\Delta K=K_f-K_0$

The final kinetic energy is:

$K_f=\frac{1}{2}(m_1+m_2)v^2_f$

We use the sum of the masses because the objects are stuck together after the collision. The initial kinetic energy is:

$K_0=\frac{1}{2}m_1v^2_{01}+\frac{1}{2}m_2v^2_{02}$

Substituting we get:

$\Delta K=\frac{1}{2}(m_1+m_2)v^2_f-(\frac{1}{2}m_1v^2_{01}+\frac{1}{2}m_2v^2_{02})$

The only missing variable is the final velocity. To determine the final velocity we will use the conservation of momentum.

We will use the conservation of momentum in the horizontal direction (west) and the conservation of momentum in the vertical direction (north).

In the horizontal direction we have:

$m_1v_{h1}+m_2v_{h2}=(m_1+m_2)v_{hf}$

Since the object 1 has no velocity in the horizontal direction we have that:

$\begin{gathered} m_1(0)+m_2v_{h2}=(m_1+m_2)v_{hf} \\ m_2v_{h2}=(m_1+m_2)v_{hf} \end{gathered}$

Now we solve for the final horizontal velocity:

$\frac{m_2v_{h2}}{\mleft(m_1+m_2\mright)}=v_{hf}$

Now we substitute the values:

$\frac{(6kg)(2\frac{m}{s})}{(4kg+6kg)}=v_{hf}$

Solving the operations we get:

$1.2\frac{m}{s}=v_{hf}$

Now we use the conservation of momentum in the vertical direction, we get:

$m_1v_{v1}+m_2v_{v2}=(m_1+m_2)v_{vf}$

Since the second object has no vertical velocity we get:

$m_1v_{v1}=(m_1+m_2)v_{vf}$

Now w solve for the final vertical velocity, we get:

$\frac{m_1v_{v1}}{\mleft(m_1+m_2\mright)}=v_{vf}$

Now we substitute the values:

$\frac{(4kg)(5\frac{m}{s})}{(4kg+6kg)}=v_{vf}$

Now we solve the operations:

$2\frac{m}{s}=v_{vf}$

Now we determine the magnitude of the final velocity using the following formula:

$v_f=\sqrt[]{v^2_{hf}+v^2_{vf}}$

Substituting the values:

$v_f=\sqrt[]{(1.2\frac{m}{s})^2+(2\frac{m}{s})^2}$

Solving the operations:

$v_f=2.53\frac{m}{s}$

Now we substitute this in the formula for the kinetic energy and we get:

$\Delta K=\frac{1}{2}(4kg+6kg)(2.53\frac{m}{s})^2-(\frac{1}{2}(4kg)(5\frac{m}{s})^2+\frac{1}{2}(6kg)(2\frac{m}{s})^2)$

Solving the operations:

$\Delta K=32J-62J=-30J$

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iren2701 [21]

Explanation:

<em>(a) On the dots below, which represent the sphere, draw and label the forces (not components) that are exerted on the sphere at point A and at point B, respectively. Each force must be represented by a distinct arrow starting on and pointing away from the dot.</em>

At point A, there are three forces acting on the sphere: weight force mg pulling down, normal force N pushing left, and static friction force Fs pushing down.

At point B, there are three forces acting on the sphere: weight force mg pulling down, normal force N pushing down, and static friction force Fs pushing right.

<em>(b) i. Derive an expression for the speed of the sphere at point A.</em>

Energy is conserved:

PE = PE + KE + RE

mgH = mgR + ½mv² + ½Iω²

mgH = mgR + ½mv² + ½(⅖mr²)(v/r)²

mgH = mgR + ½mv² + ⅕mv²

gH = gR + ⁷/₁₀ v²

v² = 10g(H−R)/7

v = √(10g(H−R)/7)

<em>ii. Derive an expression for the normal force the track exerts on the sphere at point A.</em>

Sum of forces in the radial (-x) direction:

∑F = ma

N = mv²/R

N = m (10g(H−R)/7) / R

N = 10mg(H−R)/(7R)

<em>(c) Calculate the ratio of the rotational kinetic energy to the translational kinetic energy of the sphere at point A.</em>

RE / KE

= (½Iω²) / (½mv²)

= ½(⅖mr²)(v/r)² / (½mv²)

= (⅕mv²) / (½mv²)

= ⅕ / ½

= ⅖

<em>(d) The minimum release height necessary for the sphere to travel around the loop and not lose contact with the loop at point B is Hmin. The sphere is replaced with a hoop of the same mass and radius. Will the value of Hmin increase, decrease, or stay the same? Justify your answer.</em>

When the sphere or hoop just begins to lose contact with the loop at point B, the normal force is 0. Sum of forces in the radial (-y) direction:

∑F = ma

mg = mv²/R

gR = v²

Applying conservation of energy:

PE = PE + KE + RE

mgH = mg(2R) + ½mv² + ½Iω²

mgH = 2mgR + ½mv² + ½(kmr²)(v/r)²

mgH = 2mgR + ½mv² + ½kmv²

gH = 2gR + ½v² + ½kv²

gH = 2gR + ½v² (1 + k)

Substituting for v²:

gH = 2gR + ½(gR) (1 + k)

H = 2R + ½R (1 + k)

H = ½R (4 + 1 + k)

H = ½R (5 + k)

For a sphere, k = 2/5. For a hoop, k = 1. As k increases, H increases.

<em>(e) The sphere is again released from a known height H and eventually leaves the track at point C, which is a height R above the bottom of the loop, as shown in the figure above. The track makes an angle of θ above the horizontal at point C. Express your answer in part (e) in terms of m, r, H, R, θ, and physical constants, as appropriate. Calculate the maximum height above the bottom of the loop that the sphere will reach.</em>

C is at the same height as A, so we can use our answer from part (b) to write an equation for the initial velocity at C.

v₀ = √(10g(H−R)/7)

The vertical component of this initial velocity is v₀ sin θ. At the maximum height, the vertical velocity is 0. During this time, the sphere is in free fall. The maximum height reached is therefore:

v² = v₀² + 2aΔx

0² = (√(10g(H−R)/7) sin θ)² + 2(-g)(h − R)

0 = 10g(H−R)/7 sin²θ − 2g(h − R)

2g(h − R) = 10g(H−R)/7 sin²θ

h − R = 5(H−R)/7 sin²θ

h = R + ⁵/₇(H−R)sin²θ

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