# How do I do this physics problem?A 70kg box is pulled by a 400N force at an angle 30 degrees to the horizontal. The force of kinetic friction is 75N. What is the acceleration of the box????

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### 3 Answers

The 70 kg box is being pulled by a 400 N force that makes an angle of 30 degrees to the horizontal. The force of kinetic friction is 75 N. The acceleration of the box has to be determined.

The force of 400 N is applied at an angle of 30 degrees to the horizontal. The horizontal component of the force is 400*cos 30 and the vertical component is 400*sin 30 = 200

There is an opposing force due to friction that decreases the horizontal component to 400*cos 30 - 75 = 271.41 N. The acceleration of the box in the horizontal direction is 271.41/70 = 3.8773 m/s^2 and the acceleration in the vertical direction is 200/70 = 2.8571 m/s^2.

The combined acceleration has a magnitude of sqrt(3.8773^2 + 2.8571^2) = 4.8162 m/s^2 at an angle 36.38 degrees to the horizontal.

The box is accelerated at 4.8162 m/s^2 at an angle of 36.38 degrees to the horizontal.

Mass of Box = 70 Kg

Pulling Force applied = 400 N

Force of Kinetic Friction = 75 N

Angle of incline = 30 degrees

Gravitational Force component along incline = 75*9.8*sin(30) = 343 N

* If the box is pulled up the incline, then effctive resisting force = 343+75 = 418 N which is greater than the pulling force of 400 N and the box will not move up the plane*.

However, if the box is pulled down the plane, then the resultant force on the box = 400+343-75 = 668 N

Acceleration = Force/Mass = 668/70 = 9.54 m/s/s

**If the box is pull down the incline, then the acceleration is 9.54 m/s/s**

**The above answer by justaguide is correct for the hirzontal acceleration of 3.8773m/s^2 and that is the only acceleration of the box.**

200 N vertical component of the force in upwards direction is less than the gravitational pull of 70*9.8 N in the downwards direction hence there will be no acceleration in the vertical direction.

**In case, this downward gravitational force is less than the vertical component of the force (upwards), then in that case there will be no frictional force in the horizontal direction and the box will fly with the resultant acceleration in a direction determined by the net horizontal and vertical accelerations.**