Answer:
Explanation:
A 97 kg cart is pushed down the hall. The cart starts from rest. The cart travels 27.634 m in 12.33 sec. Since not much is mentioned about the pushing force, it is assumed constant.
So, the cart is under a uniformly accelerated linear motion.
Also, since nothing is mentioned about friction, it is assumed that the surface is smooth.
Consider the equation of motion, [tex]S=ut+\frac{1}{2}at^{2}[/tex]
Here [tex]S[/tex] is the distance covered, [tex]u[/tex] is the initial velocity, [tex]a[/tex] is the acceleration, [tex]t[/tex] is the time.
On substituting the values,
[tex]27.634=(0)(12.33)+\frac{1}{2}a(12.33)^{2}\\a=0.364 \frac{m}{s^{2}}[/tex]
So, force applied = [tex]m\times a=97\times0.364=35.26\text{ }N[/tex]
∴ Force applied on the cart = 35.26 N
