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A vertical rope AB has its end B attached to the top of a scale pan - Edexcel - A-Level Maths Mechanics - Question 2 - 2016 - Paper 1

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A vertical rope AB has its end B attached to the top of a scale pan. The scale pan has mass 0.5 kg and carries a brick of mass 1.5 kg, as shown in Figure 1. The scal... show full transcript

Worked Solution & Example Answer:A vertical rope AB has its end B attached to the top of a scale pan - Edexcel - A-Level Maths Mechanics - Question 2 - 2016 - Paper 1

Step 1

Find the tension in the rope AB.

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Answer

To find the tension in the rope AB, we need to consider the forces acting on the system. The total mass being accelerated is the sum of the mass of the scale pan and the mass of the brick:

mtotal=0.5extkg+1.5extkg=2.0extkgm_{total} = 0.5 ext{ kg} + 1.5 ext{ kg} = 2.0 ext{ kg}

The weight of the entire system is given by:

W=mtotalimesg=2.0imes9.81extm/s2=19.62extNW = m_{total} imes g = 2.0 imes 9.81 ext{ m/s}² = 19.62 ext{ N}

Since the system is accelerating upwards with an acceleration of 0.5 m/s², we apply Newton's second law:

TW=mtotalimesaT - W = m_{total} imes a

Replacing W:

T19.62=2.0imes0.5T - 19.62 = 2.0 imes 0.5

T=19.62+1.0=20.62extNT = 19.62 + 1.0 = 20.62 ext{ N}

Thus, the tension in the rope AB is approximately 20.6 N.

Step 2

Find the magnitude of the force exerted on the scale pan by the brick.

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Answer

To find the magnitude of the force exerted on the scale pan by the brick, we consider only the brick's mass and the forces acting on it. The weight of the brick is:

Wbrick=1.5extkgimes9.81extm/s2=14.715extNW_{brick} = 1.5 ext{ kg} imes 9.81 ext{ m/s}² = 14.715 ext{ N}

The brick also experiences an upward force due to the tension in the rope, so we can apply Newton's second law to the brick:

TbrickWbrick=mbrickimesaT_{brick} - W_{brick} = m_{brick} imes a

Substituting the known values:

Tbrick14.715=1.5imes0.5T_{brick} - 14.715 = 1.5 imes 0.5

Tbrick14.715=0.75T_{brick} - 14.715 = 0.75

Solving for T_{brick} gives:

Tbrick=14.715+0.75=15.465extNT_{brick} = 14.715 + 0.75 = 15.465 ext{ N}

Therefore, the magnitude of the force exerted on the scale pan by the brick is approximately 15.5 N.

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