if a block of mass 2kg slides on an inclined plane making an angle of 60 with the horizontal.then what will be the work done by the force of gravity ?

Question

IN THE GIVEN FIG ...THERE IS A RIGHT ANGLED TRIANGLE IN WHICH HEIGHT =3M ,BASE =4M .AND THE OBEJECT IS PLACED AT A TOP OF TRIANGLE SLIDE OVER HYPOTENUSE MAKING ANGLE 60 WITH BASE ....DETAIL OF GIVEN FIG ASIDE

Answer 1

The problem statement has an error.  If the height is 3m and the base is 4m, the angle must be tan^-1(3/4) = 36.87°.  You can see it is not 60°.  I will assume the base and height are correct.  In any case, you will see the method for solving the problem.

 

There are 3 ways to do this:

 

(1)

 

If there is no friction while sliding down the plane, the potential energy at the top is fully converted to kinetic energy at the bottom.  If the block started from rest, then the change in kinetic energy is the kinetic energy at the bottom.  Work is the change in kinetic energy, which in this problem, equals the potential energy at the top.

 

W = (1/2)mv² = mgh = (2 kg)(9.8 m/s²)(3m) = 58.8 J

 

(2)

 

Again, if no friction, then work equals minus the change in potential energy.

 

W = -ΔU = -(U_f - U_i) = -[mg(0) - mgh] = mgh = (2 kg)(9.8 m/s²)(3m) = 58.8 J

 

(3)

 

Project the weight of the block in the direction of travel, which is parallel to the surface of the plane, and use

 

work = (force along direction of travel) x (distance traveled)

 

Parallel to the surface of the plane, the force of gravity is

 

mg_x = mg sin(36.87º)  [This may be hard to see without a picture.]

 

The distance traveled is the length of the hypotenuse:  d = √(4² + 3²) = 5m

 

W = [mg sin(36.87º)] d = (2 kg)(9.8 m/s²) sin(36.87º) (5m) = 58.8 J

 

So you have 3 ways of solving, all giving the same answer.  If the base and heights are different, just replace them in the solutions.  Could you email to me a copy of the diagram that came with this problem?