Question

A 0.200-kg mass attached to the end of a spring causes it to stretch 5.0 cm. If another 0.200-kg mass is added to the spring, the potential energy of the spring will be (show work, of explain your answer) A) twice as much. B) the same. C) 4 times as much. D) one-half as much. E) 3 times as much. Pushing on the pump of a soap dispenser compresses a small spring. When the spring is compassed 0.50 cm, its potential energy is 0.0025 J. What is the spring constant of the spring? (Show work) A diver who weighs 500 N steps off a diving board that is 10 m above the water. The diver hits the water with kinetic energy of A) 5000 J. B) 510 J. C) 500 J. D) 10 J. E) more than 5000 J. A 40.0-kg suitcase is being pulled along long an inclined ramp at an airport, by means of a strap which exerts a force of 10.0 N at an angle of 47.0 degree above the horizontal. If the ramp makes an angle of 15 degree above the horizontal. How much work is done by this force on the suitcade when it move 10.0 m along the ramp? How work or explain your answer.
A) 84.8 J
B) 68.2 J
C) 100 J
D) 96.6 J
E) 46.6 J

Answer 1

Answer:

A: 4 times as much

B: 200 N/m

C: 5000 N

D: 84,8 J

Explanation:

A.

In the first question, we have to caculate the constant of the spring with this equation:

$m*g=k*x$

Getting the k:

$k=\frac{m*g}{x} =\frac{0,2[kg]*9,81[\frac{m}{s^{2} } ]}{0,05[m]} =39,24[\frac{N}{m}]$

Then we can calculate how much the spring stretch whith the another mass of 0,2kg:

$x=\frac{m*g}{k} =\frac{0,4[kg]*9,81[\frac{m}{s^{2} } ]}{39,24[\frac{N}{m}]} =0,1[m]$

The energy of a spring:

$E=\frac{1}{2}*k*x^{2}$

For the first case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,05[m])^{2} =0,049 [J]$

For the second case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,1[m])^{2} =0,0196 [J]$

If you take the relation E2/E1 = 4.

B.

We have the next facts:

x=0,005 m

E = 0,0025 J

Using the energy equation for a spring:

$E=\frac{1}{2}*k*x^{2}$⇒$k=\frac{E*2}{x^{2} } =\frac{0,0025[J]*2}{(0,005[m])^{2} } =200[\frac{N}{m} ]$

C.

The potential energy of the diver will be equal to the kinetic energy in the moment befover hitting the watter.

$E=W*h=500[N]*10[m]=5000[J]$

Watch out the units in this case, the 500 N reffer to the weighs of the diver almost relative to the earth, thats equal to m*g.

D.

The work is equal to the force acting in the direction of the motion. so we have to do the diference beetwen angles to obtain the effective angle where the force is acting: 47-15=32 degree.

The force acting in the direction of the ramp will be the projection of the force in the ramp, equal to F*cos(32). The work will be:

W=F*d=F*cos(32)*d=10N*cos(32)*10m=84,8J