A force of $50 N$ is inclined to the vertical at an angle of $30^{o}$ . Find the acceleration it produces in a body of mass $2 k g$ which moves in the horizontal direction.

Open in App

Solution

cae I
next horizontal force is $F sin 30^{o}$
$a = \frac{F o r c e}{m a s s} = \frac{50 sin 30^{o}}{2} = 12.5 m / s^{2}$
case II
next horizontal force $= F sin 30^{o}$
$a = \frac{F o r c e}{m a s s} = \frac{50 sin 30^{o}}{2} = 12.5 m / s^{2}$
Only friction will differ in the two cases

Suggest Corrections

Similar questions

Q. A force of

10 N

inclined to the horizontal at an angle of

60^{o}

acts on a body of mass

2 K g

. If the body can move in horizontal direction only find its acceleration.

Q. A force of

72

dyne is inclined to the horizontal at an angle of

60^{o}

. Find the acceleration it produces in a mass of

9 g

which moves in a horizontal direction.

A force of $50 N$ is applied on a body of mass $2 k g$ at an angle of 30° with the vertical. Find the acceleration with which the body moves in the horizontal direction.

Q. A force of 64 dyne acts on a body inclined to the horizontal at an angle of

60^{o}

. Find the acceleration in the body if its mass is 8 g and it can only move in the horizontal direction.

Q. A force of

10

N inclined to the horizontal at an angle of

60

acts on a body can move in horizontal direction only find its acceleration.

Join BYJU'S Learning Program

A force of 50 N is inclined to the vertical at an angle of 30o. Find the acceleration it produces in a body of mass 2 kg which moves in the horizontal direction.

cae Inext horizontal force is Fsin30oa=Forcemass=50sin30o2=12.5m/s2case IInext horizontal force =Fsin30oa=Forcemass=50sin30o2=12.5m/s2Only friction will differ in the two cases

A force of $50 N$ is inclined to the vertical at an angle of $30^{o}$ . Find the acceleration it produces in a body of mass $2 k g$ which moves in the horizontal direction.

cae I
next horizontal force is $F sin 30^{o}$
$a = \frac{F o r c e}{m a s s} = \frac{50 sin 30^{o}}{2} = 12.5 m / s^{2}$
case II
next horizontal force $= F sin 30^{o}$
$a = \frac{F o r c e}{m a s s} = \frac{50 sin 30^{o}}{2} = 12.5 m / s^{2}$
Only friction will differ in the two cases