Question 3

Block A is placed on a rough surface inclined at an angle θ above the horizontal. A taut string connects block A over a pulley to block B, which hangs from the string, as shown below. The masses of blocks A and B are MA and MB, respectively. At time t=0, block A is sliding up the slope as block B falls, and the blocks are both slowing down. Assume that the mass and friction of the pulley are negligible.
If the mass of block B is 2kg, the gravitational force exerted on block B is most nearly which of the following?

Newton's 2nd law explains that the acceleration produced by the resultant force on an object is proportional and in line with the resultant force and inversely proportional to the mass of the object

∑F = m. a

[tex] \large {\boxed {\bold {a = \frac {\sum F} {m}}} [/tex]

F = force, N

m = mass = kg

a = acceleration m / s²

In this 2-beam system the forces acting on both:

Beam A = mg sin θ and string tension (T)

Beam B = mg and string tension (T)

So the force acting on the system:

∑F = m. a

mg sin θ -T + T-mg = m.a

Because what is being asked is the gravitational force acting on the block B , then :

F = mg

F = 2 kg. 10 m / s²

F = 20 N

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Keywords: system acceleration, The masses of blocks,friction, the gravitational force

abidemiokin abidemiokin · Answer 2 · 2021-10-26T15:37:36+02:00

The gravitational force exerted on block B is most nearly 20N

According to Newton's second law of motion;

[tex]\sum F = ma[/tex]

[tex]\sum F[/tex] is the sum of force acting on the block

m is the mass of the block B

a is the acceleration of the block

The acceleration on the block B is the acceleration due to gravity i.e. a = g

F = mg

Given the following parameters

m = 2.0kg

g = 9.8m/s²

Substitute the given parameters into the formula

F = 2 × 9.8

F = 19.6N

F ≈ 20 N

Hence the gravitational force exerted on block B is most nearly 20N

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