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PilotLPTM [1.2K]
3 years ago
14

Find an expression for the kinetic energy of the car at the top of the loop.Express the kinetic energy in terms of m, g, h, and

R.
Physics
1 answer:
lyudmila [28]3 years ago
3 0

Answer:

K.E₂ = mg(h - 2R)

Explanation:

The diagram of the car at the top of the loop is given below. Considering the initial position of the car and the final position as the top of the loop. We apply law of conservation of energy:

K.E₁ + P.E₁ = K.E₂ + P.E₂

where,

K.E₁ = Initial Kinetic Energy = (1/2)mv² = (1/2)m(0 m/s)² = 0 (car initially at rest)

P.E₁ = Initial Potential Energy = mgh

K.E₂ = Final Kinetic Energy at the top of the loop = ?

P.E₂ = Final Potential Energy = mg(2R) (since, the height at top of loop is 2R)

Therefore,

0 + mgh = K.E₂ + mg(2R)

<u>K.E₂ = mg(h - 2R)</u>

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What is the difference between a bar magnet and an electromagnet? (check all that are
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D and A

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A roller coaster cart of mass m = 223 kg starts stationary at point A, where h1 = 26.8 m and a while later is at B, were h2 = 14
Tresset [83]

Answer:

vB = 15.4 m/s

Explanation:

Principle of conservation of energy:

Because there is no friction the mechanical energy is conserve

ΔE = 0

ΔE : mechanical energy change (J)

K : Kinetic energy (J)

U: Potential energy (J)

K = (1/2)mv²

U = m*g*h

Where :

m: mass (kg)

v : speed (m/s)

h : hight (m)

Ef - Ei = 0

(K+U)final - (K+U)initial =0

(K+U)final = (K+U)initial

((1/2)mv²+m*g*h)final = ((1/2)mv²+m*g*h)initial , We divided by m both sides of the equation:

((1/2)vB² + g*hB = (1/2 )vA²+ g*hA

(1/2) (vB)² + (9.8)*(14.7) =  0 + (9.8)(26.8 )

(1/2) (vB)² = (9.8)(26.8 ) - (9.8)*(14.7)

(vB)² = (2)(9.8)(26.8 - 14.7)

(vB)² = 237.16

v_{B} = \sqrt{237.16}

vB = 15.4 m/s : speed of the cart at B

4 0
3 years ago
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