Ashopper in a supermarket pushes a cart with a force of 48.0 n directed at an angle of 25.0° below the horizontal. the force is just sufficient to balance various friction forces, so the cart moves at a constant speed.

a. find the work done by the shopper as she moves down a 50.0 m length aisle.
b. what is the net work done on the cart?
c. the shopper goes down the next aisle, pushing horizontally and maintaining the same speed as before. if the work done by frictional forces doesn't change, would the shopper's applied force be larger, smaller , or the same? what about the work done on the cart by the shopper?

a)

b)

c) force required would be lesser, Work done will be constant

Explanation:

Given:

force,

angle of force,

(a)

displacement,

As the displacement is horizontal, we find the fomponent of the force in the direction of displacement.

∴Work done:

(b)

Net work done on the cart is same as the work done by the shopper.

(c)

While pushing horizontally the force required would be lesser because it is all in the direction of displacement.

Work done will be constant because the force in the direction of a constant displacement would be the same.

  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

it would be kevlar ^^