Question

A 30 kg block is to be moved up an inclined plane at an angle 30° to the horizontal with a velocity of 5 m/s. If the frictional force retarding the motion is 150N. Find the horizontal force required to move the block up the plane.(g = 10m/s²)​

Answer 1

Refer the attachment

F = 300N

From the above figure, Force of friction n = 150N

Net horizontal force,

F - f = mg sin

F - mg sin 30° + f

30 × 10 ×

1

2

+150

is the horizontal force required to move up

Answer 2

$\huge\bf\underline{Answer}:-$

From the figure (attachment),

• Force of Friction = 150 N

Net Horizontal Force,

$\sf F-f=mg\sin\theta$

$\\ \implies\sf F=mg\sin30^°+f$

$\\ \implies\sf F=30\times10\times\dfrac{1}{2}+150$

$\\ \implies\sf F=30\times\cancel{10}\ \ ^5\times\dfrac{1}{\not{2}}+150$

$\\ \implies\sf F=30\times5+150$

$\\ \implies\sf F=150+150$

$\\ \therefore\sf F=300\ N.$

The horizontal force required to move the block up the plane is 300 N.