Answer:
(a) 4.75 m/s² (b) 7.6 N
Explanation:
Given that,
The mass of a lab cart, m = 1.6 kg
Initial speed, u = 0.5 m/s
Final speed, v = 4.3 m/s
Time, t = 0.8 s
(a) The acceleration of an object is equal to the rate of change of velocity. It can be calculated as :
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{4.3-0.5}{0.8}\\\\a=4.75\ m/s^2[/tex]
(b) Let F be the net force on the cart. It can be given by :
F = ma
[tex]F=1.6\times 4.75\\\\F=7.6\ N[/tex]
Hence, this is the required solution.
