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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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A block of mass m = 150 kg rests against a spring with a spring constant of k = 880 N/m on an inclined plane which makes an angl
weeeeeb [17]

Answer:

b)  k Δx - W cos θ - μ mg cos θ = m a ,  c)  θ = 86.6º, d)  Δx = 1.18 m

Explanation:

a) In the attachment we can see a diagram of the forces in this problem and the coordinate axes for its decomposition.

F is the force applied by the spring, while it is compressed, this force disappears when the block leaves the spring

b) Let's apply Newton's second law for when the spring is compressed

let's use trigonometry to break down the weight

            sin θ = Wₓ / W

            cos θ = W_y / W

             Wₓ = W sin θ

             W_y = W cos θ

Y axis  

               N - W_y = 0

               N = W_y

               N = W cos θ

X axis

           F -Wₓ -fr = ma

the force applied by the spring is given by hooke's law

           F = k Δx

friction force has the expression

           fr = μ N

           fr = μ W cos θ

we substitute

            k Δx - W cos θ - μ mg cos θ = m a           ( 1)

c) If the plane has no friction, what is the angle so that Δx = 0.1m

             

We write the equation 1, with fr = 0 and since the system is still a = 0

            k Δx - W cos θ -0 = 0

            cos θ = \frac{k \Delta x}{ m g}

            cos θ = \frac{880 \ 0.1}{ 150 \ 9.8}

            cos θ = 0.0598

            θ = cos⁻¹ 0.0598

            θ = 86.6º

d) In this part they give the angle θ = 45º and there is no friction, they ask the compression

the acceleration is zero, we substitute in 1

            k Δx - W cos θ - 0 = 0

            Δx = \frac{mg \ cos \  \theta}{k}

            Δx = \frac{ 150 \ 9.8 \ cos45}{880}

            Δx = 1.18 m

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To increase the validity of the results obtained from the single experiment, the students should be encouraged to repeat the experiments more number of times as much as possible.

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A single observation/experiment is not valid enough.

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