Question

Figure 3
(b) The surface is tilted to an angle of 37 degrees from the horizontal, as shown above in Figure 3 The blocks are each given a push so that at the instant show
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they are
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1. Each block has a mass of 0.20 kg Calculate the magnitude of the net force exerted on the two block system at the instant shown above

Answer 1

Answer:

  3.92 N (straight down)

  2.36 N (down, parallel to the plane)

Explanation:

The only force acting on the system is apparently that due to gravity. The mass of the system is the sum of the masses of the blocks, so is ...

  0.20 kg + 0.20 kg = 0.40 kg

The force due to gravity is given by ...

  F = ma = (0.40 kg)(9.8 m/s²) = 3.92 N . . . (straight down)

That force is acting straight down, so is at an angle with respect to the direction the blocks are constrained to move.

The force down the plane is (3.92 N)·sin(37°) ≈ 2.359 N . . . (down the plane)

_____

Comment on the answer

We have given two answers to the question, because in this frictionless system, the force acting normal to the plane of motion is irrelevant to the system dynamics. The question asked for the net force on the system, which is the force due to gravity, so we have given that magnitude also.

Answer 2

The parallel force on the block system is 2.36 N

The perpendicular force on the bock system is 3.13 N.

The given parameters;

• mass of the first block, m₁ = 0.2 kg

• mass of the second block, m₂ = 0.2 kg

• angle of inclination, θ = 37⁰

The magnitude of the net force exerted on the block system is calculated as follows;

The parallel force on the block system is calculated as;

$F_{||} = mg \times sin(\theta)\\\\F_{||} = (0.2 + 0.2)\times 9.8 \times sin(37)\\\\F_{||} = 2.36 \ N$

The perpendicular force on the bock system;

F⊥ = mg x cos(Ф)

F⊥ = (0.2 + 0.2) x 9.8 x cos(37)

F⊥ = 3.13 N

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