A block of mass 3.54kg is pushed 1.19m along a frictionless horizontal table by a constant 16N force directed 27 degrees below the horizontalA.)...

Given:

Mass of block = m = 3.54 kg

Distance moved = s = 1.19 m

Force applied on the block = f1 = 16 N

Angle (below horizontal) of application of force f1 = A = 27 degree

A) Work done by the applied force:

Since there is movement only in the horizontal direction, work is done only by the horizontal component of force (f).

f = f1*(Cos A) = 16*(Cos 27) = 16*0.8910 = 14.256

Further, as there is no friction the application of the force f will accelerated the block. This acceleration (a) of block is given by formula:

a = f/m = 14.256/3.54 = 4.0271

In absence of any frictional resistance the work done by this force is entirely converted to kinetic energy e. This is given by the formula:

e = m*a*s = 3.54*4.0271*1.19 = 16.96464 J

The energy could also gave been calculated directly as:

e = f*s = 14.256*1.19 = 16.96464 J

B) Work done by the normal force exerted by table:

As there is no movement in a direction normal to the surface of the table, the work done by normal force exerted by the table is 0.

C) Work done by the force of gravity:

As there is no movement in vertical direction the work done by force of gravity is 0.

D) Work done by the net force on the block:

The only movement caused by the net force is the horizontal movement of the block. Therefore the total work done by the net force on the block is same as the applied force, as calculated in A) above.