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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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...
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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.
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
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