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lana [24]
3 years ago
13

A 15 kg runaway grocery cart runs into a spring with spring constant 230 N/m and compresses it by 56 cm .What was the speed of t

he cart just before it hit the spring?
Physics
1 answer:
liubo4ka [24]3 years ago
3 0

To solve this problem we will apply the concepts related to the conservation of kinetic energy and elastic potential energy. Thus we will have that the kinetic energy is

KE = \frac{1}{2} mv^2

And the potential energy is

PE = \frac{1}{2} kx^2

Here,

m = mass

v = Velocity

x = Displacement

k = Spring constant

There is equilibrium, then,

KE = PE

\frac{1}{2} mv^2 = \frac{1}{2} kx^2

Our values are given as,

x=0.56m\\k=230N/m\\m=15kg

Replacing we have that

\frac{1}{2} (15)v^2 = \frac{1}{2} (230)(0.56)^2

v = 2.19m/s

Therefore the speed of the cart is 2.19m/s

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Answer:

1030.83\ \text{N}

Explanation:

\Delta L = Cambio en la longitud de la cuerda = 0.25 cm

T = tensión en cuerda

A = Área de la cadena = \dfrac{\pi}{4}d^2

d = Diámetro de la cuerda = 0.2 cm

L = Longitud original de la cuerda = 1.6 m

El cambio de longitud de una cuerda viene dado por

\Delta L=\dfrac{TL}{AE}\\\Rightarrow T=\dfrac{\Delta LAE}{L}\\\Rightarrow T=\dfrac{0.25\times 10^{-2}\times \dfrac{\pi}{4}(0.2\times 10^{-2})^2\times 210\times 10^9}{1.6}\\\Rightarrow T=1030.84\ \text{N}

La tensión en la cuerda es 1030.84\ \text{N}.

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