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A body of mass 40kg resting on a rough h...

A body of mass 40kg resting on a rough horizontal surface is subjected to a force `P` which is just enough to start the motion of the body. If `mu_(s)=0.5mu_(k)=0.4 , g=10ms^(-2)` an dthe force `P` is continuously applied on the body, then the accceleration of the body is.

A

a. zero

B

b. `1 m//s^(2)`

C

c.`2 m//s^(2)`

D

d. `2.4 m//s^(2)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we need to analyze the forces acting on the body and apply Newton's laws of motion. ### Step 1: Identify the forces acting on the body The forces acting on the body are: - The applied force \( P \) - The static friction force \( f_s \) (which is at its maximum when the body is just about to move) - The weight of the body \( W = mg \) Given: - Mass \( m = 40 \, \text{kg} \) - Coefficient of static friction \( \mu_s = 0.5 \) - Coefficient of kinetic friction \( \mu_k = 0.4 \) - Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \) ### Step 2: Calculate the maximum static friction force The maximum static friction force can be calculated using the formula: \[ f_s = \mu_s \cdot N \] where \( N \) is the normal force. On a horizontal surface, the normal force \( N \) is equal to the weight of the body: \[ N = mg = 40 \, \text{kg} \times 10 \, \text{m/s}^2 = 400 \, \text{N} \] Now, substituting the values: \[ f_s = \mu_s \cdot mg = 0.5 \cdot 400 \, \text{N} = 200 \, \text{N} \] ### Step 3: Determine the force \( P \) when the body is just about to move When the body is just about to start moving, the applied force \( P \) is equal to the maximum static friction force: \[ P = f_s = 200 \, \text{N} \] ### Step 4: Calculate the kinetic friction force once the body starts moving Once the body starts moving, the frictional force acting on it is the kinetic friction force \( f_k \): \[ f_k = \mu_k \cdot N = \mu_k \cdot mg = 0.4 \cdot 400 \, \text{N} = 160 \, \text{N} \] ### Step 5: Apply Newton's second law to find the acceleration According to Newton's second law: \[ F_{\text{net}} = P - f_k = ma \] Substituting the known values: \[ 200 \, \text{N} - 160 \, \text{N} = 40 \, \text{kg} \cdot a \] \[ 40 \, \text{N} = 40 \, \text{kg} \cdot a \] Dividing both sides by 40 kg: \[ a = 1 \, \text{m/s}^2 \] ### Final Answer The acceleration of the body is \( 1 \, \text{m/s}^2 \). ---
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