A 4.2x10^3-N piano is to be slid up a 3.5-m frictionless plank at a constant speed. The plank makes an angle of 30 deg. with the horizontalcalcwI need help with physics question

2 Answers

krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on


Force exerted by the piano = f = 4.2x10^3 N

Angle of the plank = A = 30 degrees

Distance to be moved = d = 3.5 m

The weight of piano, and their the force exerted by it is acting in vertically downward direction.

The component of gravitational force of piano acting along the plank (f1) is given by formula:

f1 = fx(Sign A)

= (4.2x10^3)x(Sin 30) = (4.2x10^3)x0.5 = 2.1x10^3

Since the the plank is frictionless there is no frictional force to be overcome to move the piano up the plank.

Also the plank is moved up at constant speed. Therefor, there is no work done to accelerate the piano when it is moved.

Therefor, the only force required to move the piano up the plank is the force requires to exactly over come the force f1 pulling the piano down the plank. Magnitude of this force is also equal to f1.

And the work done to move the piano is given by:

Work done = (Force)x(Distance moved) = (f1)x(d)

= 2.1x10^3x3.5 = 7.35x10^3 N-m

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

The weight force F of the piano is  4.2*10^3 where g is the acceleration due to gravity and its value is 9.81m/s^2.

The distance s covered in sliding the piano = 3.5m.

Therefore the work done by the piano insliding = Force component along the plank * distance covered = Fsin30*S= 4.2*10^3N sin 30 *3.5 = 7.35*10^3Nm