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A 15.0 kg cart is moving with a velocity of 7.50 m/s down a level hallway. A constant force of 10.0 N acts on the cart, and its velocity become 3.20 m/s.
a) What is the change in kinetic energy?
b) How much work was done on the cart?
c) How far did the cart move while the force acted? ...?

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Below are the answers:

a. DE=421.875-76.8=-345.075 joules (negitive sign means that the energy is transfering out of your system ie slowing down) 

b. work=DE (work done is change in energy)=-345.075 joules 

c. W=f*d 

d=w/f 

d=345.075/10 

d=34.5075 meters
Eduard22sly

A. The change in kinetic energy is –345.075 J

B. The amount of work done is –345.075 J

C. The cart travelled a distance of 34.51 m

A. How to determine the change in kinetic energy

  • Mass (m) = 15 Kg
  • Initial velocity (u) = 7.5 m/s
  • Final velocity (v) = 3.20 m/s
  • Change in kinetic energy (ΔKE) =?

ΔKE = ½m(v² – u²)

ΔKE = ½ × 15 × (3.2² – 7.5²)

ΔKE = 7.5 × –46.01

ΔKE = –345.075 J

B. How to determine the workdone

The workdone in this case is equal to the change in energy of the cart.

  • Change in kinetic energy (ΔKE) = –345.075 J
  • Workdone = Change in kinetic energy (ΔKE)
  • Workdone = –345.075 J

C. How to determine the distance

  • Workdone (Wd) = –345.075 J
  • Force (F) = –10 N
  • Distance (d) =?

Wd = Fd

Divide both side by F

d = Wd / F

d = –345.075 / –10

d = 34.51 m

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