A 15.5 kg block is pulled by two forces. The first is 11.8N at a 53.7 angle and the second is 22.9 N at a -15.8 angle. What is the direction of the acceleration

α = 6.43° (Angle respect to the horizontal axis-x)

Explanation:

To solve this problem we must use Newton's second law which tells us that the sum of forces must be equal to the product of mass by acceleration.

Since we have two forces and we know their magnitudes and their directions, we can sum these into their X & y components, in this way we can find the resulting force and apply it in Newton's second law.

F1x = 11.8*cos(53.7) = 6.98 [N]

F1y = 11.8*sin(53.7) = 9.5 [N]

Now with the second force:

F2x = 22.9*cos (15.8) = 22.03 [N]

F2y = - 22.9*sin (15.8) = - 6.23 [N]

Now we sum the forces in the x and y axes:

Fx = F1x + F2x = 6.98 + 22.03 = 29.01 [N]

Fy = F1y + F2y = 9.5 -  6.23 = 3.27 [N]

Now using the Pythagorean theorem we can find the resulting force.

F = √(Fx² + Fy²)

F = √ (29.01)² + (3.27)²

F = 29.19 [N]

Using Newton's second law, we have:

F = m*a

where:

F = force = 29.19 [N]

m = mass = 15.5 [kg]

a = acceleration [m/s²]

a = 29.19/15.5

a = 1.88 [m/s²]

The direction of the acceleration is the same direction of the force, therefore we need to find the angle.

tan(α) = Fy/Fx

tan(α) = 3.27/29.01

α = tan⁻¹(0.1127)

α = 6.43° (Angle respect to the horizontal axis-x)

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer

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