Question

Desperate for help! please!!!!!!!!

as seen from the side, a 3.74 kg box on a flat frictionless ground is pulled forward by a 4.20 n force at a 50.0 degree angle and pulled back by a 2.25 N force at a 122 degree angle. What is the acceleration of the box?

Answer 1

0.4029 m/s2

Step-by-step explanation:

Like for the Acellus Gang!

Answer 2

`0.4029 \mathrm{m} / \mathrm{s}^(2)` is the acceleration of the box.

Step-by-step explanation:

Given data:

Mass of the box = 3.74 kg

Flat friction-less ground is pulled forward by a 4.20 N force at a 50.0 degree angle and pulled back by a 2.25 N force at a 122 degree angle.

First, we need to find the net horizontal force acting on the box. With the given data, the equation can be formed as below. Net horizontal force acting on the box (F) is given by

`F=\left(4.20 * \cos 50^(\circ)\right)+\left(2.25 * \cos 122^(\circ)\right)`

`F=(4.20 * 0.64278)+(2.25 *-0.5299)`

F = 2.699676 – 1.192275 = 1.507 N

Next, find acceleration of the box using Newton's second law of motion. This states that the link between mass (m) of an objects and the force (F) required to accelerate it. The equation can be given as

`F=m * acceleration\ (a)`

`\text {acceleration }(a)=(F)/(m)=(1.507)/(3.74)=0.4029 \mathrm{m} / \mathrm{s}^(2)`