3. Yumyum pulls a box to the left with 300 N of force. Mamu also pulls the box to the left with 250 N of force. The box has a mass of 25 kg.
a. What is the net force acting on the box?
b. What is the box's acceleration?

The net force acting on the box is 550 N (going leftwards) and the box's acceleration is 22 meters per second squared.

Explanation:

A)

The net force acting on the box is the sum of the forces. Since Yumyum pulls with a force of 300 N and Mamu pulls with a force of 250 N, both to the left, the net force (in the x-direction) will be:

B)

The net force is equal to the object's mass multiplied by its acceleration. That is:

Substitute known values:

Find a:

In conclusion, the net force acting on the box is 550 N (going leftwards) and the box's acceleration is 22 meters per second squared.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s