Atoy jeep rolls down an incline at a constant velocity. the jeep's mass is m = 1.7 kg and the ramp makes an angle of θ = 12 degrees with respect to the horizontal. assume the rolling resistance is negligible. part (a) what is the magnitude of the jeep's acceleration, a in m/s2? numeric : a numeric value is expected and not an expression. a = part (b) what is the numeric value for the sum of the forces in the x-direction, σfx, in newtons?

a=0

∑Fx = 0

Explanation:

Known data

m =1.7 kg   :  jeep's mass

θ = 12°  :angle of  the slope with respect to the horizontal direction

g = 9.8 m/s² : acceleration due to gravity

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the block on the ramp and the y-axis in the direction perpendicular to it.

Calculated of the W

W= m*g

W=  1.7 kg* 9.8 m/s² = 16.66 N

x-y weight components

Wx= Wsin θ= 16.66*sin 12° = 3.45 N

Wy= Wcos θ =16.66*cos 12° = 16.3 N

a) Because the toy jeep lowers the ramp at a constant speed, then, its acceleration is equal to zero.

a=0

b)We apply the formula (1) to calculated  ΣFx

∑Fx = m*a  , a=0

∑Fx = 0

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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