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A body of mass 40 kg stands on a weighin...

A body of mass 40 kg stands on a weighing machine in an accelerated lift. The reading on the scale of the weighing machine is 300 N. Find the magnitude and direction of acceleration. `(g = 9.8ms^(-2))`

A

2.3 `m//s^2` upward

B

2.3 `m//s^2` downward

C

4.6 `ms//^2` upward

D

4.6 `ms//^2` downward

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the forces acting on the body when it is in an accelerated lift. Here’s a step-by-step breakdown: ### Step 1: Identify the Forces When the body is in the lift, two main forces act on it: 1. The gravitational force (weight) acting downwards, which is given by: \[ W = m \cdot g \] where \( m = 40 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \). 2. The normal force (N) exerted by the weighing machine, which is the reading on the scale. ### Step 2: Calculate the Weight of the Body Using the formula for weight: \[ W = 40 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 392 \, \text{N} \] ### Step 3: Analyze the Situation in the Lift Since the reading on the scale is 300 N, we can set up the equation based on Newton's second law. If the lift is accelerating downwards, the normal force will be less than the weight: \[ N = W - F_{net} \] where \( F_{net} = m \cdot a \) (the net force due to the acceleration of the lift). ### Step 4: Set Up the Equation From the above, we can write: \[ N = m \cdot g - m \cdot a \] Substituting the known values: \[ 300 \, \text{N} = 392 \, \text{N} - 40 \, \text{kg} \cdot a \] ### Step 5: Solve for Acceleration Rearranging the equation to solve for \( a \): \[ 40 \, \text{kg} \cdot a = 392 \, \text{N} - 300 \, \text{N} \] \[ 40 \, \text{kg} \cdot a = 92 \, \text{N} \] \[ a = \frac{92 \, \text{N}}{40 \, \text{kg}} = 2.3 \, \text{m/s}^2 \] ### Step 6: Determine the Direction of Acceleration Since the normal force (300 N) is less than the weight (392 N), the lift is accelerating downwards. Therefore, the direction of acceleration is downward. ### Final Answer The magnitude of acceleration is \( 2.3 \, \text{m/s}^2 \) and it is directed downwards. ---
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