a person moves a full sam\'s club shopping cart mass=80.0 kg from rest to a velo

Question

a person moves a full sam's club shopping cart mass=80.0 kg from rest to a velocity of 1.30 m/s over a distance of 1.80 meters to secure a spot in check out line.

A. find the time required to get the cart moving from rest to 1.30 m/s?

B. Determine the change in the chart's momentum?

c. determine the magnitude of the force exerted on the cart by the person.

D. the result from part c ignores friction. if the shopping cart's coefficient of friction is .40 determine the force required by the person to move the cart. (recall the friction force = UFn)

Explanation / Answer

a) avg velocity = (final - initial)/2 = (1.30 - 0)/2 = 0.65

time = distance / avg velocity = 1.80 / 0.65 = 2.77 seconds

b) initial momentum = 0

   final momentum = mv = 80 * 1.30 =   104 N-s

change in momentum = 104 N-s

c)   force = change in momentum / time = 104 / 2.77 = 37.56 Newtons

d) force of friction = umg = 0.4*80*9.8 = 313.6 Newtons

    force of person = orginal force + friction = 37.56 + 313.6 =   351.16 Newtons

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