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# A spring with a force constant of k...?

I need A LOT of help on physics homework

A spring with a force constant of k=500N/m is compressed horizontally by a 10kg mass along a rough surface. The spring is compressed 0.3m, and the coefficient of kinetic friction between the mass and the surface is 0.15. If the system is released, what will be the approximate velocity of the mass when the spring is no longer compressed?

An engine maintains constant power on a conveyor belt machine. If the belt's velocity is doubled, the magnitude of its average acceleration

a. is doubled

b. is quartered

c. is halved

d. is quadrupled

e. remains the same

A box is pulled along a smooth floor by a force F, making an angle (theta) with the horizontal. As (theta) increases, the amount of work done to pull the box the same distance, d,

a. increases

b. increases and then decreases

c. remains the same

d. decreases and then increases

e. decreases

As the time needed to run up a flight of stairs decreases, the amount of work done against gravity

a. increases

b. decreases

c. remains the same

d. increases and then decreases

e. decreases and then increases

A box of mass m slides down a frictionless inclined plane of length L, and vertical height h. What is the change in its gravitational potential energy, PE?

a. -mgL

b. -mgh

c. -mgL/h

d. -mgh/L

e. -mg/L

Could you explain please? I'd like to be able to understand the homework so I can do well on the test as well

### 2 Answers

- Anonymous8 years agoFavorite Answer
1)

Energy stored in spring = Es = 1/2*k*u^2 (u = distance compressed)

Energy in a moving mass = Em = 1/2*m*v^2 (m is mass, v is velocity)

Energy dissipated by friction = Ef = m*g*c*u (g is gravity constant 9.8 m/s^2, c is coefficient of kin. friction)

Es = Em + Ef

Em = Es - Ef

1/2*m*v^2 = 1/2*k*u^2 - m*g*c*u

v = sqrt( k*u^2/m - 2*g*c*u )

v = sqrt( 500*0.3^2/10 - 2*9.8*0.15*0.3 ) = 1.9 m/s

2)

Ill defined question in my opinion, twice as much speed would mean twice as much power required from the motor to keep the belt running. The remaining power can be translated to left over force to accelerate the belt, Premaining = (Pconstant - Pvelocity), however this does not relate to any of the answers given in the multiple choice.

3)

a) increases, the componet of gravity against which you pull is m*g*sin(theta). This becomes larger for larger theta. You could say it decreases again for theta > 90 degrees, but then the floor would be upside down and the box falls off.

4)

c) Amount of work = energy to run up. You still run up the same height, so work stays the same. What does increase is the power required, but since this higher power is multiplied by a lower time, still the work stays the same.

5)

b) -m*g*h, because you have to look at the distance traveled parallel to gravity, which is h. L has nothing to do with it.

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- brechtelLv 43 years ago
EDITED reaction the priority is what's meant with the aid of "two times as far"? that could desire to propose halve the gap which factors ?s = 0.2cm, or it may desire to propose yet another 0.4 cm. it is likewise no longer sparkling regardless of if the size given is the gap from uncompressed length or the compressed length itself. The means saved in a (linear) spring is E = 0.5*ok*s^2; ?E is the replace in means, ?s is the replace in compression. in this occasion, ?s is one 0.5 of 0.40 = 0.20 cm. The means replace is 0.0046J. From the above equation ok = 2*?E/(s1^2 - s2^2) the conception that the size indicated is distance from uncompressed length is the only one that comes on the component to your answer.

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