Block A of mass 35 kg is resting on a frictionless floor. Another block B of mass 7 kg is resting on it as shown in figure. The coefficient of static friction between the blocks is 0.5, while coefficient of kinetic friction is 0.4. If a force of 100 N is applied to block B, acceleration of block A will be (Take g = 10 `m s^2`)

Options

A. `0.8 ms^(-2)`
B. `2.4 ms^(-2)`
C. `0.4 ms^(-2)`
D. `4.4 ms^(-2)`

Answer

A

Explanation

Here `m_A = 35 kg m_b = 7 kg , mu_k = 0.5 , mu_k = 0.4 `
` F= 100 N , g= 10 ms^(-2)`
static frictional force on B
`F_s = mu_s m_B g = 0.5 xx 7 xx 10 =35 N`
As ` F gt F_s ` therefore A and B will move in the same direction i.e., of applied force , but with different accelerations
Dynamic frictiional force on B
` f_k = mu_k m_B = 0.4 xx 7 xx 10 = 28 N`
this will oppose the motion of B and cause the motion of A . for B , the equation of motion is
` F - f_K = m_B a_B`
`therefore 100 -28 = 7a_B or a_B = ( 72 )/(7) =10.3 ms^(-2)`
for A , the equation of motion in ` m_A a_A = f_k`
` therefore 35 a_A = 28 or a_A ( 28)/( 35 ) = 0.8 ms^(-2)`