# 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.) Find the work done by the applied force. Answer in...

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 horizontal

A.) Find the work done by the applied force. Answer in units of J.

B.) Find the work done by the normal force exerted by the table. Answer in units of J.

C.) Find the work done by the force of gravity. Answer in units of J.

D.) Find the work done by the net force on the block. Answer in units of J.

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

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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.

neela | High School Teacher | (Level 3) Valedictorian

Posted on

The force exerted = 16N directed 27 deg nelow  the horizontal.

a)

Therefore the work done by the force = distance * force component along the horizontal = (1.19meter)* 16N*cos 27deg=16.9648 J

b)

Work done by the mormal force = component of normal force* distance in the direction of normal force =(16 sin 27)*0 = 0. Since there is no movement of the block along the direction of normal force.

c)

Force of gravity =( mass of the block ) * gravitational atraction = 3.54*g N

Work done by 3.54g = (3.54g)* displacement along the gravitational force = (3.54g) * 0 = 0 J, as there is no vertical motion.

d)

Since there is no frictional force ,the work done by the net force on the block = The sum of the work done as above = 16.9648 J + 0 + 0 .