university physics problem X5

[匿名]

2009-09-25 3:00 pm

1.) a 1.8-kg block is moved at constant speed over a surface for which "mu"k=0.25. displacement is 2 m. it is pulled bya force directed at 45 degree to the horiaontal.
find a)the force F b)friction c)gravity

2.)a 200-g ball thrown vertically up with an initial speed of 20m/s reaches the miximum height of 18m. find a)the change of kinteic energy b)the work done by gravity c) Are two qantities just calculated equal?

3.)a 2-Kg block is projected at 3m/s up at a 15 degree incline for which "mu"k=0.2. use the work energy theorem to find the distance it travels before coming rest.

4.)a locust can propel its body(~3g)from rest to 3.4m/s in 4 cm . estimate the average power output of its legs.

5.)A 2-kg block travels on a tabletop in a circle of radius 1.2m. in one revolution its speed drops from 8 to 6m/s how many more revolutions does it make before stopping?

please show all the works and include all the fromula

回答 (1)

Prof. Physics

2009-09-25 5:27 pm

✔ 最佳答案

1.a. Friction = ukmg = (0.25)(1.8)(10) = 4.5 N

Fcos45* = 4.5

Force, F = 6.36 N

c. Gravity, g = 10 Nkg-1

2. At maximum height, the velocity is zero.

Change in K.E. = 1/2 mv2

= 1/2 (0.2)(20)2

= 40 J

b. Work done by gravity

= G.P.E. gained

= mgh

= (0.2)(10)(18)

= 36 J

c. The two quantities are not equal, as some of the energy may lose to the surroundings as heat, or overcoming air resistance.

3. By Work-energy theorem,

Loss in K.E. = Gain in G.P.E. + Work done against friction

1/2 mv2 = mglsin@ + (ukmgcos@)l

1/2 (3)2 = (10)sin15* l + 0.2(10)cos15* l

Distance travelled, l = 0.996 m

4. By v2 = u2 + 2as

(3.4)2 = 0 + 2a(0.04)

Average acceleration, a = 144.5 ms-2

By v = u + at

3.4 = 0 + 144.5t

Time, t = 0.0235 s

By Newton's 2nd law of motion,

F = ma

= (0.003)(144.5)

= 0.4335 N

So, the average power output

= Fs / t

= 0.4335 X 0.004 / 0.0235

= 0.0737 W

5. Loss in K.E. = 1/2 m(v2 - u2) = 1/2 (2)(82 - 62) = 28 J

To become at rest, loss in K.E. = 1/2 mv2 = 1/2 (2)(6)2 = 36 J

So, the required distance to stop

= 36/28 X (2pi X 1.2)

= 9.694 m

= 1.54 rev

參考: Physics king

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