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cupoosta [38]
2 years ago
9

Derive an expression for the total mechanical energy of the system as the monkey reaches the top of the motion, Etop, in terms o

f m, x, d, k, the maximum height above the bottom of the motion, hmax, and the variables available in the palette.
Physics
1 answer:
ipn [44]2 years ago
8 0

Answer:

U =  0.5 * k *(x + d - h_max)^2 + m*g*h_max

Explanation:

Given:

- The extension in spring @ equilibrium = x m

- The spring constant = k

- The amount of distance pulled down = d

- mass of the toy = m

Find:

- The total mechanical energy E_top at the top position h_max in terms of the available variables.

Solution:

- First we need to determine the types of Energy that are in play:

- The Elastic potential Energy E_p in a spring is given:

                              E_p: 0.5 * k * (ext)

- In our case when the toy at the top most position h_max will have a net extension ext, by summing displacement of spring:

             ext = Equilibrium + distance pulled - h_max = (x + d - h_max)

Hence, the elastic potential energy will be:

                              E_p = 0.5 * k *(x + d - h_max)^2

- The gravitational potential energy E_g is given by:

                              E_g = m*g*h_max

Where, bottom most position is taken as reference (datum).

- The kinetic Energy E_k is given by:

                              E_k = 0.5*m*v_top^2

- Since we know that the maximum height is reached when velocity is zero

Hence,                   E_k = 0.5*m*0^2 = 0.

The total Energy of the system U is sum of all energies and play:

                               U = E_p + E_k + E_g

                               U =  0.5 * k *(x + d - h_max)^2 + m*g*h_max

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shtirl [24]

Q: ken, 0.75 kg, moves toward a wall (his path normal to the wall) at 52 m/s. 13.0 ms after he touches the wall he pushes himself off in the opposite direction at 60 m/s. What is the magnitude of the average force the wall exerts on Ken during this rapid maneuver

Answer:

-6461.54 N

Explanation:

From Newton's Fundamental equation,

F = m(v-u)/t.................... Equation 1

Where F = Force exerted in sonic, m = mass of ken, v = final velocity, u = initial velocity, t = time.

Given: m = 0.75 kg, v = - 60 m/s (opposite direction), u = 52 m/s, t = 13 ms = 0.013 s

Substitute into equation 1

F = 0.75(-60-52)/0.013

F = 0.75(-112)/0.013

F = -84/0.013

F = -6461.54 N

Note: The negative sign tells that the force act in opposite direction to the initial motion of ken.

Hence the magnitude of the average force of the wall = -6461.54 N

4 0
2 years ago
A Honda Civic and an 18 wheeler approach a right angle intersection and then collide. After the collision, they become interlock
Goryan [66]

Answer:

The concept of conservation of momentum is applied in the particular case of collisions.  

The general equation ig given by,

M_1V_1 + M_2V_2 = (M_1+M_2) * V_f,

Where,

M_2 = 4 M_1

The crash occurs at an intersection so we must separate the two speeds by their respective vector: x, y.

In the case of the X axis, we have that the body M_2 has a speed = 0, this because it is not the direction in which it travels, therefore

M_1* 13 = (M_1+M_2) * V_{fx} \\M_1*13 = (M_1+4M_1)*V_{fx}\\M_1*13=5M_1*V_{fx}\\Vx = \frac{13}{5}m/s

The same analysis must be given for the particular case in the Y direction, where the mass body M_1 does not act with its velocity here, therefore:

M_2* 13 = (M_1+M_2) * V_{fy},\\4*M_1* 13 = 5Ma * V_{fy} ,\\V_{fy} = \frac{52}{5}m/s ,

We have the two components of a velocity vector given by V_f = \frac{13}{5}\hat{i} + \frac{52}{5}\hat{j}

Get the magnitude,

V_f = \sqrt{(\frac{13}{5})^2+(\frac{52}{5})^2}

V_f = 10.72 m/s

With a direction given by

Tan^{-1} \frac{4}{1} = 75.96 \°

8 0
2 years ago
Megan rode the bus to school, which is located 8 kilometers from her home. If Megan's frame of reference is her house, and it to
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Answer:

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2 years ago
A cart starts from rest and accelerates at 4.0 m/s2 for 5.0 s, then maintains that velocity for 10 s, and then decelerates at th
zhannawk [14.2K]

Answer:

Final speed of car = 12 m/s

Explanation:

We have equation of motion v = u + at, where v is final velocity, u is initial velocity, a is acceleration and t is time.

a) A cart starts from rest and accelerates at 4.0 m/s² for 5.0 s

        v = ?

        u = 0 m/s

        a = 4.0 m/s²

         t = 5 s

         v = u + at = 0 + 4 x 5 = 20 m/s

b) Then maintains that velocity for 10 s

        v = ?

        u = 20 m/s

        a = 0 m/s²

         t = 10 s

         v = u + at = 20 + 0 x 10 = 20 m/s

c) Then decelerates at the rate of 2.0 m/s² for 4.0 s

        v = ?

        u = 20 m/s

        a = -2.0 m/s²

         t = 4 s

         v = u + at = 20 + -2 x 4 = 12 m/s

Final speed of car = 12 m/s

3 0
2 years ago
A 1500 kg car traveling at 20 m/s suddenly runs out of gas while approaching the valley shown in the figure. The alert driver im
geniusboy [140]

Answer:

v_f = 17.4 m / s

Explanation:

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final point. Arriving at the gas station

         Em_f = K + U = ½ m v_f ² + m g y₂

energy is conserved

         Em₀ = Em_f

         ½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂

        v_f ² = v₀² + 2g (y₁ -y₂)

         

we calculate

        v_f ² = 20² + 2 9.8  (10 -15)

        v_f = √302

         v_f = 17.4 m / s

8 0
2 years ago
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