Two forces are applied to a block of mass MM=12.0 kg that sits on a horizontal, frictionless table. One force F⃗ 1F→1 has a magnitude of 10.0 N and is directed at an angle θ1 equal to 37.0° from the xx direction as labeled in the figure below. As a result of the forces, the block experiences an acceleration a equal to 0.440 m/s2 in the xx direction only. What is the x component of the sum of the other forces acting on the block? (Express your answer to three significant figures.) 2)
What is the yy component of the sum of the other forces acting on the block? (Express your answer to three significant figures.)
I cant figure out Part 2. HELP
To find the x component of the sum of the other forces acting on the block, we first need to find the x component of the force F1 and then subtract it from the total force.
1. Calculate the x component of the force F1:
F1x = F1 * cos(θ1)
F1x = 10.0 N * cos(37.0°)
F1x ≈ 8.00 N
2. Subtract the x component of F1 from the total force:
F2x = Total force - F1x
F2x = M * a - F1x
F2x = 12.0 kg * 0.440 m/s² - 8.00 N
F2x ≈ 3.92 N
Therefore, the x component of the sum of the other forces acting on the block is approximately 3.92 N.
To find the y component of the sum of the other forces acting on the block, we need to find the y component of the force F1 and then add it to the total force.
3. Calculate the y component of the force F1:
F1y = F1 * sin(θ1)
F1y = 10.0 N * sin(37.0°)
F1y ≈ 6.00 N
4. Add the y component of F1 to the total force:
F2y = Total force + F1y
F2y = M * a + F1y
F2y = 12.0 kg * 0.440 m/s² + 6.00 N
F2y ≈ 11.28 N
Therefore, the y component of the sum of the other forces acting on the block is approximately 11.28 N.
Part 1
F = 12*0.44 = 5.28 N
Therefore,
10cos37 + F2 = 5.28
F2 = -2.71 N