A block of mass 2 kg slides on an inclined plane which makes an angle of 30° with the horizontal. The coefficient of friction between the block and the surface is √(3/2) . What force along the plane should be applied to the block so that it moves down?

Question

A block of mass 2 kg slides on an inclined plane which makes an angle of 30° with the horizontal. The coefficient of friction between the block and the surface is √(3/2) . What force along the plane should be applied to the block so that it moves down? (A) 11.21 (B) 4.5 (C) 6.7 (D) 30.14

Tardigrade Admin · Accepted Answer

mgsinθ =(.2.)(.10.) (. (1/2) .) =10N=F1 (say) Solution

μ mg cosθ = (. √(3/2) .) (.2.)(.10.) (. (√3/2) .) =21.21N=F2 (say) Force required to move the block down the plane with constant velocity. Solution

F 1 will be acting downwards, while F 2 upwards. Since F2>F1, force required F=F2-F1=11.21 N

Q. A block of mass $2 k g$ slides on an inclined plane which makes an angle of $3 0^{\circ}$ with the horizontal. The coefficient of friction between the block and the surface is $\frac{3}{2}$ . What force along the plane should be applied to the block so that it moves down?

$11.21$

$4.5$

$6.7$

$30.14$

Q. A block of mass 2kg slides on an inclined plane which makes an angle of 30∘ with the horizontal. The coefficient of friction between the block and the surface is 23​​ . What force along the plane should be applied to the block so that it moves down?

11.21

4.5

6.7

30.14

mgsinθ=(2)(10) (21​) =10N=F1​ (say) μmgcosθ= (23​​) (2)(10) (23​​) =21.21N=F2​ (say) Force required to move the block down the plane with constant velocity. F1​ will be acting downwards, while F2​ upwards. Since F2​>F1​, force required F=F2​−F1​=11.21N

Q. A block of mass $2 k g$ slides on an inclined plane which makes an angle of $3 0^{\circ}$ with the horizontal. The coefficient of friction between the block and the surface is $\frac{3}{2}$ . What force along the plane should be applied to the block so that it moves down?

$11.21$

$4.5$

$6.7$

$30.14$