Question

A person pulls a box on a horizontal floor by a rope making an angle 30 to the normal the tension force on the rope is 400 N and the friction force is 45 N, find the acceleration of the box if its weight is 1292 N.

a)1.5 m/s²
b)5 m/s²
c)0 m/s²
d) non of them
I need the answer urgently please ​

Answer 1

Answer: First, let's resolve the forces acting on the box along the horizontal axis:

T cos(30) - f = ma

where T is the tension force, f is the friction force, m is the mass of the box, and a is its acceleration.

We can also resolve the forces along the vertical axis:

T sin(30) - W = 0

where W is the weight of the box.

Solving for T and W:

T = 400 N

W = 1292 N

T sin(30) = (400 N) sin(30) = 200 N

W = 1292 N

Now we can substitute these values into the horizontal equation and solve for the acceleration:

T cos(30) - f = ma

(400 N) cos(30) - (45 N) = (m)a

(346.4 N) = (m)a

a = (346.4 N) / m

We can calculate the mass of the box using its weight:

W = mg

1292 N = m(9.81 m/s^2)

m = 131.8 kg

Now we can substitute the mass into the equation:

a = (346.4 N) / (131.8 kg)

a ≈ 2.63 m/s^2

Therefore, the correct answer is not listed. The acceleration of the box is approximately 2.63 m/s^2.

Step-by-step explanation: