Physics, 26.07.2019 15:30, aupein

A10 kg runaway grocery cart runs into a spring with spring constant 250 n/m and compresses it by 60 cm. what was the speed of the cart just before it hit the spring?

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answered: 99keevintaylor012

The initial speed of the cart is 3 m/s.

What is the velocity?

The velocity of an object is the rate of change of its position with respect to a given time interval.

Given that the mass m of the cart is 10 kg and the spring constant is 250 N/m. The compressed length of the spring is 60 cm.

The initial kinetic energy of the cart is given below.

$KE = \dfrac {1}{2}mv^2$

Where v is the initial velocity of the cart.

When the cart hits the spring, it is compressed by the cart. When the cart stops, due to the maximum compression, the kinetic energy of the cart will be zero. Hence its kinetic energy will be converted into elastic potential energy of the spring. This is given as,

$U = \dfrac {1}{2} kx^2$

Where k is the spring constant and x is the compressed length of the spring.

$U = \dfrac {1}{2} \times 250 \times 0.60^2$

$U = 45 \;\rm J$

This elastic potential energy of the spring will be equal to the initial kinetic energy of the cart.

$45 = \dfrac {1}{2} \times 10 \times v^2$

90 = 10 v^2

v^2 = 9

$v = 3 \;\rm m/s$

Hence we can conclude that the initial speed of the cart is 3 m/s.

To know more about the velocity, follow the link given below.

link.

answered: spazzinchicago

The initial kinetic energy of the cart is
$K= \frac{1}{2} mv^2$ (1)
where m is the mass of the cart and v its initial velocity.

Then, the cart hits the spring compressing it. The maximum compression occurs when the cart stops, and at that point the kinetic energy of the cart is zero, so all its initial kinetic energy has been converted into elastic potential energy of the spring:
$U= \frac{1}{2}kx^2$
where k is the spring constant and x is the spring compression.

For energy conservation, K=U. We can calculate U first: the compression of the spring is x=60 cm=0.60 m, while the spring constant is k=250 N/m, so
$U= \frac{1}{2}kx^2= \frac{1}{2}(250 N/m)(0.60 m)^2=45 J$

So, the initial kinetic energy of the cart is also 45 J, and from (1) we can find the value of the initial velocity:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 45 J}{10 kg} } =3 m/s$

answered: Guest

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6 $\frac{m}{s^{2} }$

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

answered: Guest

Wooden , walls ,doors ,support

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A10 kg runaway grocery cart runs into a spring with spring constant 250 n/m and compresses it by 60...