Question

You push a 1.00 food tray through the cafeteria One

Answer 1

Answer:

Acceleration of tray and carton = 6.03 m/s²

Horizontal force exerted by tray in carton = 3.02 N

Explanation:

Given:

Mass of tray (M) = 1.00 kg

Mass of milk carton (m) = 0.50 kg

Force applied on the system (F) = 9.04 N

The tray and carton slide horizontally without friction.

The free body diagram of the system is shown below.

The net force acting on the tray and carton system is the constant force 9.04 N.

The total mass of the system is (M + m) = 1.00 kg + 0.50 kg = 1.50 kg

Now, according to Newton's second law:

Net Force = mass × acceleration

⇒ Acceleration = Net force ÷ mass

Therefore, the acceleration of the tray and carton is given as:

$a=\frac{9.04}{1.50}=6.03\ m/s^2$

Now, consider the free body diagram of the carton alone.

The forces acting on the carton is only the horizontal force (f) exerted by the tray.

So, from Newton's second law:

Net Force = mass × acceleration

$f=ma\\f=0.50\times 6.03\\f=3.02\ N$

Therefore, the horizontal force exerted by the tray on the carton is 3.02 N.