dETERMINE the horizontal force needed to accelerate a 50 kg grocery cart from rest to 0.90 m/s in 2.8s if the coefficient of friction between the cart and the floor is 0.20?

a = (.90 -0)/2.8

weight = m g = 50 (9.81)

friction force = .2 * 50 (9.81)

F - .2*50(9.81) = 50 (.9)/2.8

did the sign * is for multplication?

yes * means "times"

To determine the horizontal force needed to accelerate the grocery cart, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.

Given:
Mass of the grocery cart (m) = 50 kg
Initial velocity (u) = 0 m/s (rest)
Final velocity (v) = 0.90 m/s
Time taken (t) = 2.8 s
Coefficient of friction (μ) = 0.20

First, let's find the acceleration of the cart using the formula:

Acceleration (a) = (v - u) / t

Substituting the given values:
a = (0.90 m/s - 0 m/s) / 2.8 s
a = 0.90 m/s / 2.8 s
a ≈ 0.321 m/s²

Now, let's find the force of friction acting on the cart using the equation:

Force of friction (Ff) = μ * (mass * gravitational acceleration)

Gravitational acceleration (g) is approximately 9.8 m/s².

Ff = 0.20 * (50 kg * 9.8 m/s²)
Ff = 0.20 * (490 N)
Ff = 98 N

Since the force of friction acts in the opposite direction of the applied force, we need to consider it as a negative value.

Therefore, the net force required to accelerate the cart is:

Net force (Fnet) = mass * acceleration + Force of friction
Fnet = (50 kg * 0.321 m/s²) + (-98 N)
Fnet = 16.05 N + (-98 N)
Fnet ≈ -81.95 N

The negative sign indicates that the force required is in the opposite direction (opposite to the applied force).

Hence, a horizontal force of approximately 81.95 N is needed to accelerate the 50 kg grocery cart from rest to 0.90 m/s in 2.8 seconds with a coefficient of friction of 0.20.