Calculate the speed for the following situations Hint a. A shopper pushes a 22-k

Question

Calculate the speed for the following situations Hint a. A shopper pushes a 22-kg shopping cart over 2.7 meters, pushing with a force of 12 newtons. Fill in the blanks below joules, and if the shopping cart rolls frictionlessly, The work that the shopper does is 32.4 its speed is ms at the end of the push. b. In the above scenario, it turns out that the shopping cart does not roll frictionlessly, but instead there is a friction force of 7.5 newtons. What is the speed of the cart at the end of the push? Hint for (b) The speed of the cart is m/s at the end of the push. c. A new scenario: the shopping cart rolls down a ramp starting from rest, and as it rolls down, the gravitational force does 27 joules of work on the cart. What is the speed of the cart at the bottom of the ramp, if you can ignore friction? The speed of the cart at the bottom of the ramp is m's. decimal number [more Enter an integer or

Explanation / Answer

a)

work done = force * displacement

= 12 N * 2.7 m

= 32.4 J

use:

work done = increase in kinetic energy

32.4 = 0.5*m*v^2

32.4 = 0.5*22*v^2

v^2 = 2.945

v = 1.72 m/s

Answer: 1.72 m/s

b)

frictional work = 7.5 N * 2.7 m

= 20.25 J

actual work = 32.4 J - frictional work

= 32.4 J - 20.25 J

= 12.15 J

use:

work done = increase in kinetic energy

12.15 = 0.5*m*v^2

12.15 = 0.5*22*v^2

v^2 = 1.1045

v = 1.05 m/s

Answer: 1.05 m/s

c)

work done = increase in kinetic energy

27 = 0.5*m*v^2

27 = 0.5*22*v^2

v^2 = 2.4545

v = 1.57 m/s

Answer: 1.57 m/s

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