Question

A 15.0kg block is dragged over a rough, horizontal surface by a 70.0-N force acting at 20.0 degrees above the horizontal. The block is displaced 5.0 m, and the coefficient of kinetic friction is 0.3.
a. Draw a free body diagram. Draw your coordinate system and label the axes.
b. Calculate the work done on the block by the 70 N force.
c. Calculate the work done on the block by the normal force.
d. Calculate the work done on the block by the gravitational force.
e. Calculate the work done on the block by the force of friction.

Answer 1

Answer:

$W=70 * 5cos20 = 328.89 J$

$W_n = 0$

$W_g=0$

$W_f= -184.59J$

Work done is 0

Explanation:

From the question we are told that

Weight of block =15.0kg

Force acting on the block = 70.0N

At an angle of 20 degree

Displacement of block is 5m

Coefficient of kinetic friction 0.3

b) Generally work done by force is give by $W=fdcos \theta$

therefore

      $W=70 * 5cos20 = 328.89 J$

c) there is no work done by the normal force in this scenario because

normal force in this case is perpendicular to the displacement of the motion

       $W_n = 0$

d) The displacement in the vertical direction is 0

Therefore the gravitational work done is 0  $W_g=0$

e)Generally in finding work done by friction we first find frictional force

Mathematically the equation for frictional force is given $f = \alpha N$

Given that

       $N=mg-Fsin20$

       $N= 15.0*9.8 - 70 sin20$

       $N=123 N$

       $f=0.3* 123.06 = 36.92N$

Mathematically solving to get work done by frictional force $W_f$

        $W_f= -fd\\W_f = -36.92 * 5$

         $W_f= -184.59J$

the frictional force work done is  $W_f= -184.59J$