please answer this....?

Question

1.Space Traveler whose mass is 75kg leaves the earth.compute his weight?
a. on earth b. on mars,where g=3.8m/s

2. A box of mass 0.6 slug is moving w/ a constant velecity of 15ft/s on a horizontal surface.The coefficient of sliding friction between the box and the surface is 0.20.what horizontal force is required to maintain the motion?

3. A dictionary is pulled to the right at consstant velocity by a 10N force acting at 30% above the horizontal.the coefficient sliding friction between the dictionary and the surface is 0.5.What is the weight of the dictionary?

Answer 1

1 a) on earth his weight is 75*9.8 N On Mars it is 75*3.8 N (that should be g(mars) = 3.8 m/sec^2)

2. The frictional force and horizontal force must be equal if the box moves at constant velocity. The frictional force is 0.6*0.2*g pounds [the weight of the object is m(slug) * g(ft/sec^2)]. g = 32.2 ft/sec^2 so the required horizontal force is 0.6*0.2*32,2 pounds.

3 The normal force from the dictionary is m*g - 10*sin(30º).
The horizontal force from friction is then 0.5*[m*g - 10*sin(30º)]. This must equal the horizontal component of the applied 10 N force, or 10*cos(30º). Therefore

0.5*[m*g - 10*sin(30º)] = 10*cos(30º)

solve for m*g (the weight).

Answer 2

1) We = weight on earth
We = mg
We = 75 kg x 9.8 m/s^2
We = 735 N ANS

Wm = weight on mars
Wm = 75 kg x 3.8 m/s^2
Wm = 285 N ANS

2) Given:
m = 0.6 slug
V = 15 ft/s
μk = 0.20

Find:
Fh = horizontal force required to maintain motion

Solution:
Fh - Ff = ma, where Ff is friction force

Fh - (μk)Fn = ma, where Fn is the normal force acting on the box
Fh - 0.20(0.6 x 32) = 0.6(0)
Fh = 0.20(0.6 x 32)
Fh = 3.84 N ANS

3)
Given:
Fa = 10 N at 30° above horizontal
μk = 0.5

Find:
Wd = weight of dictionary

Solution:
Fax - Ff = ma
Fa(cos 30°) - μk(Fn) = ma
10(cos 30°) - 0.5(Wd) = (Wd/9.8)(0)
10(cos 30°) = 0.5(Wd)
Wd = 10(cos 30°)/0.5
Wd = 17.3 N ANS


teddy boy

Answer 3

1) (a) The weight on earth is P = m g = 75*9.8 = 735 N.
(b) on mars is: P' = m g' = 75*3.8 = 285 N
2) m = 0.6 slug = 0.6*14.59 = 8.754 Kg
v = 15 ft/s = 15*0.3048 = 4.572 m/s
F = 0.2 * m g = 17.16 N
3) If 30% means 30°, then : F cos30° - 0.5 N = 0
F sin30° - mg + N = 0
N = mg - F sin30°
F (cos30° + 0.5 sin30°) = 0.5 mg
mg = 2*10(cos30° + 0.5 sin30°) = 22.32 N