Answer:
F = 51.3°
Explanation:
The component of weight parallel to the inclined plane must be responsible for the rolling back motion of the car. Hence, the force required to be applied by the child must also be equal to that component of weight:

where,
W = Weight of Wagon = 150 N
θ = Angle of Inclinition = 20°
Therefore,

<u>F = 51.3°</u>
Answer:
a = 2.72 [m/s2]
Explanation:
To solve this problem we must use the following kinematics equation:

where:
Vf = final velocity = 1200 [km/h]
Vo = initial velocity = 25 [km/h]
t = time = 2 [min] = 2/60 = 0.0333 [h]
1200 = 25 + (a*0.0333)
a = 35250.35 [km/h2]
if we convert these units to units of meters per second squared
![35250.35[\frac{km}{h^{2} }]*(\frac{1}{3600^{2} })*[\frac{h^{2} }{s^{2} } ]*(\frac{1000}{1} )*[\frac{m}{km} ] = 2.72 [\frac{m}{s^{2} } ]](https://tex.z-dn.net/?f=35250.35%5B%5Cfrac%7Bkm%7D%7Bh%5E%7B2%7D%20%7D%5D%2A%28%5Cfrac%7B1%7D%7B3600%5E%7B2%7D%20%7D%29%2A%5B%5Cfrac%7Bh%5E%7B2%7D%20%7D%7Bs%5E%7B2%7D%20%7D%20%5D%2A%28%5Cfrac%7B1000%7D%7B1%7D%20%29%2A%5B%5Cfrac%7Bm%7D%7Bkm%7D%20%5D%20%3D%202.72%20%5B%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%20%7D%20%5D)