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A boy whose mass is 50kg stands on a spring balance inside a lift. The lift starts to ascent with an acceleration of  2ms−2.2𝑚𝑠−2. The reading of the machine or balance is  (g=10 ms−2)(𝑔=10 ms−2)

Question

A boy whose mass is 50kg stands on a spring balance inside a lift. The lift starts to ascent with an acceleration of  2ms−2.2𝑚𝑠−2. The reading of the machine or balance is  (g=10 ms−2)(𝑔=10 ms−2)

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Solution

The force acting on the boy is the sum of the gravitational force and the force due to the acceleration of the lift.

  1. First, calculate the gravitational force acting on the boy. This is given by the equation F = m*g, where m is the mass of the boy and g is the acceleration due to gravity.

    F_gravity = 50kg * 10ms^-2 = 500N

  2. Next, calculate the force due to the acceleration of the lift. This is given by the equation F = m*a, where m is the mass of the boy and a is the acceleration of the lift.

    F_lift = 50kg * 2ms^-2 = 100N

  3. Add these two forces together to find the total force acting on the boy.

    F_total = F_gravity + F_lift = 500N + 100N = 600N

  4. The reading on the spring balance is equal to the total force acting on the boy divided by the acceleration due to gravity.

    Reading = F_total / g = 600N / 10ms^-2 = 60kg

So, the reading on the spring balance would be 60kg.

This problem has been solved

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