Answer:
The distance moved by the cart is 3.05 m
Explanation:
Given;
mass of the cart, m = 15.5 kg
applied force, f = 10.5 N
initial velocity, u = 0
time of motion, t = 3s
The final velocity of the cart is given by;
[tex]F = ma\\\\F = m(\frac{v-u}{t})\\\\ m(v-u) = Ft\\\\m(v-0) = Ft\\\\mv = Ft\\\\v = \frac{Ft}{m}\\\\ v = \frac{10.5*3}{15.5}\\\\ v = 2.032 \ m/s[/tex]
The acceleration of the cart is given by;
F = ma
a = F/m
a = 10.5 / 15.5
a = 0.677 m/s²
The horizontal distance moved by the cart is given by;
s = ut + ¹/₂at²
s = 0 + ¹/₂ (0.677)(3)²
s = 3.05 m
Therefore, the distance moved by the cart is 3.05 m
