Question

The diagram below shows a block of mass m = 2.0 kg on a frictionless horizontal surface, as seen from above. Three forces of magnitudes F₁ = 4.0 N, F₂ = 6.0 N, and F₁ = 8.0 N are applied to the block, initially at rest on the surface, at angles shown on the diagram. In this problem, you will determine the resultant (total) force vector from the combination of the three individual force vectors. All angles should be measured counterclockwise from the positive x axis (i.e., all angles are positive).

Part A 

Calculate the magnitude of the total resultant force F_r = F₁ + F₂ + F₃ acting on the mass.
Express the magnitude of the resultant force in newtons to to two decimal places..

Part B

What angle does F_r make with the positive x axis?
Express your answer in degrees to two significant figures.

Part C

What is the magnitude of the mass’s acceleration vector, a?
Express your answer to two significant figures.

Part D

What is the direction of a? In other words, what angle does this vector make with respect to the positive x axis?
Express your answer in degrees to two significant figures.

Part E

How far (in meters) will the mass move in 5.0 s?
Express the distance in meters to two significant figures.

Part F

What is the magnitude of the velocity vector of the block at t = 5.0 s?
Express your answer in meters per second to two significant figures.

Part G

In what direction is the mass moving at time t = 5.0 s? That is, what angle does the velocity vector make with respect to the positive x axis?
Express your answer in degrees to two significant figures.

Answer 1

Part A 

Just find the x and y components of each force and add them together:

F₁:
F_{1, x} = cos(25) * 4 = 3.625
F_{1, y} = sin(25) * 4 = 1.690

F₂:
F_{2, x} = cos(325) * 6 = 4.915
F_{2, y} = sin(325) * 6 = -3.441

F₃:
F_{3, x} = cos(180) * 8 = -8.000
F_{3, y} = sin(180) * 8 = 0.000

F_{r, x} = 3.625 + 4.915 – 8.000 = 0.54
F_{r, y} = 1.690 – 3.441 + 0.000 = -1.751

Now use the Pythagorean theorem to find the overall magnitude:

F_r = sqrt(F_{r, x}^2 + F_{r, y}^2)
F_r = sqrt(0.2916 + 3.067)
F_r = 1.83 N

1.83 N

Part B

Use the x and y components, and overall magnitude, from Part A:

F_{r, x} = 0.54
F_{r, y} = -1.751
F_r = 1.83

sin(θ) = F_{r, y} / F_r
sin(θ) = -1.751 / 1.83
sin(θ) = -0.9568
θ = -73°

-73°

Part C

F = ma

Since the force (Part A) is 1.83 N and the mass is 2.0 kg:

F = ma
1.83 = 2.0 * a
a = 0.915 m/s²

Note: Mastering Physics wants the answer rounded for this part, but you should save the significant figures for the following parts because you will need to use acceleration again – and Mastering Physics will expect you NOT to have rounded it. Mastering Physics has many problems like this, where it tells you to do one thing but expects another.

a = 0.92 m/s²

Part D

This will be the same angle as we found in Part B (-73°)

-73°

Part E

The acceleration (Part C) is 0.915 m/s² (remember that we rounded it for Part C, but need the extra significant figure here). We can use the formula below:

x = v₀t + 1/2at²
x = 0 * 5.0 + 1/2 * 0.915 * 5^2
x = 0.4575 * 5^2
x = 11 m

11 m

Part F

The acceleration (Part C) is 0.915 m/s² (remember we rounded it to 0.92 m/s² in Part C, but Mastering Physics expects the extra significant figure here). Just multiply this by 5, which gives you 4.6 m/s. However, there is another problem – Mastering Physics has ANOTHER rounding error and may expect an answer of 4.5 m/s. This is because it takes the 11 m from Part E and uses it in the following equation:

V²_f = V²_i + 2a(x).

This answers 4.5 m/s. So in case, Mastering Physics doesn’t accept your answer, this is probably why.

4.6 m/s

Part G

This is the same angle we found in Parts B and D (-73°)

-73°