A 30 kg box has to move up an inclined slope of 30° to the horizontal at a uniform velocity of 5 m/s. If the frictional force retarding the motion is 250 N the horizontal force in newtons to move up is (g=10 text m/s2) :

Question

A 30 kg box has to move up an inclined slope of 30° to the horizontal at a uniform velocity of 5 m/s. If the frictional force retarding the motion is 250 N the horizontal force in newtons to move up is (g=10 text m/s2) : (A) 400 √2 N (B) 400 N (C)  (400√3/2)N  (D)  400× 2√3N

Tardigrade Admin · Accepted Answer

A box is moving with a uniform velocity, then the force acting on the body is given by F=mg sin θ +μ mg cos θ =30× 10× sin 30o+250 =30× 10× (1/2)+250=400 Now, horizontal component of force is given by F= cos θ =400× cos 300=(400√3/2)N

Q. A 30 kg box has to move up an inclined slope of $30^{\circ}$ to the horizontal at a uniform velocity of 5 m/s. If the frictional force retarding the motion is 250 N the horizontal force in newtons to move up is $(g = 10 m / s^{2})$ :

400 $2$ N

400 N

$\frac{400 3}{2} N$

$400 \times 23 N$

Q. A 30 kg box has to move up an inclined slope of 30∘ to the horizontal at a uniform velocity of 5 m/s. If the frictional force retarding the motion is 250 N the horizontal force in newtons to move up is (g=10m/s2) :

400 2​ N

400 N

24003​​N

400×23​N

A box is moving with a uniform velocity, then the force acting on the body is given by F=mgsinθ+μmgcosθ =30×10×sin30o+250 =30×10×21​+250=400 Now, horizontal component of force is given by F=cosθ=400×cos300=24003​​N

Q. A 30 kg box has to move up an inclined slope of $30^{\circ}$ to the horizontal at a uniform velocity of 5 m/s. If the frictional force retarding the motion is 250 N the horizontal force in newtons to move up is $(g = 10 m / s^{2})$ :

400 $2$ N

$\frac{400 3}{2} N$

$400 \times 23 N$