Agirl pushes a 10.0kg sled at a constant speed by applying a force of 75n at an angle of 30 degrees with respect to the horizontal. the sled is pushed over a distance of 15 meters. how much work is done by the force of friction?

Answer

given,

mass of sled = 10 Kg

Force applied by girl = ?

angle of inclination = 30°

distance = 15 m

Sled is moving with constant speed hence acceleration will be equal to zero.

now,

F - mg sin θ - f = m a

F - mg sin θ - f = 0

F - mg sin θ = f

where f is the frictional force acting opposite to the applied force

mg sin θ is the component of the weight which is opposing the motion of the sled

now,

f = F - mg sin θ

f = 75 - 10 x 9.8 x sin 30°

f = 26 N

now,

Work done = f. s cos ∅

where ∅ is the angle between frictional  force and displacement which is equal to 180°

W= 26 x 15 x cos 180°

W = -390 J

work done by frictional force is equal to W = -390 J

he kicked it 10 feet straight up into the air? i think this works.

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz