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A man is standing on a weighing machine ...

A man is standing on a weighing machine placed in a lift. When stationary his weight is recorded as 40 kg . If the lift is accelerated upwards with an acceleration of `2m//s^(2)` , then the weight recorded in the machine will be `(g=10m//s^(2))`

A

32 kg

B

40 kg

C

42 kg

D

48 kg

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to determine the weight recorded by the weighing machine when the lift is accelerating upwards. Here are the steps to arrive at the solution: ### Step 1: Understand the Initial Condition When the lift is stationary, the weight recorded by the weighing machine is equal to the gravitational force acting on the man. This is given by: \[ W = mg \] where: - \( m = 40 \, \text{kg} \) (mass of the man) - \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity) ### Step 2: Calculate the Gravitational Force Using the values provided: \[ W = 40 \, \text{kg} \times 10 \, \text{m/s}^2 = 400 \, \text{N} \] ### Step 3: Analyze the Situation When the Lift is Accelerating Upwards When the lift accelerates upwards with an acceleration \( a = 2 \, \text{m/s}^2 \), the effective weight (normal force \( N \)) recorded by the weighing machine will change. The net force acting on the man can be expressed as: \[ N - mg = ma \] ### Step 4: Rearranging the Equation Rearranging the equation gives us: \[ N = mg + ma \] ### Step 5: Substitute the Values Now, substituting the values into the equation: - \( m = 40 \, \text{kg} \) - \( g = 10 \, \text{m/s}^2 \) - \( a = 2 \, \text{m/s}^2 \) We get: \[ N = 40 \, \text{kg} \times 10 \, \text{m/s}^2 + 40 \, \text{kg} \times 2 \, \text{m/s}^2 \] \[ N = 400 \, \text{N} + 80 \, \text{N} \] \[ N = 480 \, \text{N} \] ### Step 6: Convert the Normal Force to Mass To find the weight recorded by the weighing machine in terms of mass, we convert the normal force back to mass using: \[ \text{Mass} = \frac{N}{g} \] \[ \text{Mass} = \frac{480 \, \text{N}}{10 \, \text{m/s}^2} = 48 \, \text{kg} \] ### Final Answer Thus, the weight recorded in the machine when the lift is accelerating upwards is **48 kg**. ---
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