Physics, 22.10.2020 19:01 josetheghost6313

The horizontal surface on which the block slides is frictionless. The speed of the block before it touches the spring is 6.0 m/s. How fast is the block moving at the instant the spring has been compressed 15 cm

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The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.

Explanation:

The spring constant is 2000 newtons per meter. Let consider the spring-block system, from Principle of Energy Conservation we can represent it by the following model:

$U_{k,1}+K_{1} = U_{k,2}+K_{2}$

$K_{2} = K_{1}+(U_{k,1}-U_{k,2})$ (Eq. 1)

Where:

$K_{1}$ , $K_{2}$ - Initial and final kinetic energies of the block, measured in joules.

$U_{k,1}$ , $U_{k,2}$ - Initial and final elastic potential energy, measured in joules.

And we expand the equation above by definitions of elastic potential energy and kinetic energy:

$\frac{1}{2}\cdot m \cdot v_{2}^{2} = \frac{1}{2}\cdot m\cdot v_{1}^{2} + \frac{1}{2}\cdot k\cdot (x_{1}^{2}-x_{2}^{2})$

$v_{2} = \sqrt{v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2}) }$ (Eq. 1b)

Where:

- Mass of the block, measured in kilograms.

- Spring constant, measured in newtons per meter.

$v_{1}$ , $v_{2}$ - Initial and final velocities of the block, measured in meters per second.

$x_{1}$ , $x_{2}$ - Initial and final positions of spring, measured in meters.

If we know that $v_{1} = 6\,\frac{m}{s}$ , $k = 2000\,\frac{N}{m}$ , $m = 2\,kg$ , $x_{1} = 0\,m$ and $x_{2} = 0.15\,m$ , the final speed of the block moving at the instant the spring has been compressed is:

$v_{2} = \sqrt{\left(6\,\frac{m}{s} \right)^{2}+\left(\frac{2000\,\frac{N}{m} }{2\,kg} \right)\cdot [(0\,m)^{2}-(0.15\,m)^{2}]}$

$v_{2}\approx 3.674\,\frac{m}{s}$

The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.

The horizontal surface on which the block slides is frictionless. The speed of the block before it t

Answer from: Quest

given data:

determine the horse power (h.p) = ?

force applied to move the box(f) = 10 pounds,

box lifted from a distance (d) = 3 feet

time taken to lift (t) = 10 sec,

the horse power is defined as amount of work required to move a load. the horse power is equal to 550 foot-pound per second

1 h.p = 550 foot-pound/second

it can also be defined as the the load that need to be move a body for certain distance in a certain amount of time.

mathematically,

h.p = (force × distance)/ time

= (10 × 3)/ 10

= 30/10

= 3 foot -pound/sec

hence, the horse power is needed to move the box is 3 foot-pound/second

important note: 1 h.p = 550 foot-pound/second = 746 watts

Answer from: Quest

what are the options.

explanation:

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