In the figure shown below , a block of mass 4 kg is attached to a spring constant 24 N/m . The coefficient og friction `mu = 0.5` . If the system is initially at rest and a constant horizontal force of 50 N is applied on the block , then calculate its speed when it is moved through a distance of 0.6 m .

Explanation

In the presence of friction , both the spring force and frictional force will oppose the motion of the block , against the applied force .

Give , k = 24 N/m
`m = 4 kg , mu= 0.5 `
` F = 50 N , x = 0.6 m `
Work done by the two opposing forces
` W_(1) = -1/2 kx^(@) - mu mgx `
Work done on the spring by F , `W_(2)=Fx`
Net work done on the spring = Gain in K.E of block
`Fx - 1/2 kx^(2) - ugx = 1/2 mv^(2)`
` 50 xx 0.6 - 1/2 xx 24 xx (0.6)^(2) - 0.5 xx4 xx 9.8 xx 0.6 = 1/2 xx 4 xxv^(2) `
`= (13.92xx2)/4 = v^(2)`
`v = 2.64 ` m/s