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.

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

therefore

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

d) The displacement in the vertical direction is 0

Therefore the gravitational work done is 0  

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

Mathematically the equation for frictional force is given

Given that

Mathematically solving to get work done by frictional force

the frictional force work done is  

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂

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c) both a and b.