Acceleration????~*10 pOints...*~?

Question

can someone heLp me soLve this physics problem???

1. a box is sliding down a plane inclined at an angle of 20° from the horizontal. find the acceleration of the box, neglecting friction and air resistance.

a. 0.342m^2g

b. 0.342 m

c. 0.342 g

d. 0.342 mg

e. 0.940m^2g

f. 0.940m

g. 0.940g

h. 0.940mg

2. a box that is hurled up the inclined plane of question 1 starts up the plane with an initial velocity of 1000cm/sec. find the speed of the box moving uo the plane 3 sec later.

a. 0 cm/sec

b. -5.5 cm/sec

c. 8.9 cm/sec

d. 10.1 cm/sec

e. 42.5 m/sec

f. 63.3 cm/sec

g. 75.3 cm/sec

h. 98.1 cm/sec

i think i already know the answer in #1...mY problem is #2

thanks for the ppl that will heLp!!!^_^

Answer 1

a) You need to find the component of weight that is directed down the incline. Gravity (g) acts vertically, so split this into an accelleration down the incline and one "normal" i.e. perpendicular to the incline:

a(down) = g * sin(20) = 0.342 * g

a(norm) = g * cos(20) = 0.94 * g

The problem asks for the down the incline which is just 0.342 * g; answer c) is correct.

b) For this part just use the equation:

Vf = Vo + a * t, where

Vf = final velocity (unknown)

Vo = initial velocity = 1000 cm/s = 10 m/s (up the incline)

a = weight acting down the incline you found in a). Note this acts in the opposite direction as the initial velocity so need to use negative sign

t = time = 3 sec

substituting:

Vf = 10 - .342 * 9.8 * 3

Vf = 0 m/s; answer a) is correct.

Good Luck!

Answer 2

1......mg sin(20) = ma      a = g sin (20)       a = 3.35 m/s^2

2...... v{f} = v{i} +at

                v {f} = 1000 + (-3.35)(3)         v{f} = 989.9 m/sec