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Mary applied a force of 10.8N on a 2.6-kg tray in pushing the tray on a counter through a distance of 50 cm. If the coefficient of friction between the tray and the counter top is 0.25. Determine the work done by Mary on the tray during the 50 cm push.​

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Eduard22sly

The work done by Mary on the tray during the 50 cm push is 2.215 J

How to determine the net force

We'll begin by calculating the frictional force. This can be obtained as follow:

  • Mass (m) = 2.6 Kg
  • Coefficient of friction (μ) = 0.25
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Normal reaction (N) = mg = 2.6 × 9.8 = 25.48 N
  • Frictional force (FÕ¢) =?

Fբ = μN

Fբ = 0.25 × 25.48

FÕ¢ = 6.37 N

Finally, we shall determine the net force. This illustrated below:

  • Frictional force (FÕ¢) = 6.37
  • Force applied (F) = 10.8 N
  • Net force (Fâ‚™) =?

Fâ‚™ = F - FÕ¢

Fâ‚™ = 10.8 - 6.37

Fâ‚™ = 4.43 N

How to determine the work done

The work done by Mary can be o brained as follow:

  • Net force (Fâ‚™) = 4.43 N
  • Distance (d) = 50 cm = 50 / 100 = 0.5 m
  • Work done (Wd) =?

Wd = Fd

Wd = 4.43 × 0.5

Wd = 2.215 J

Thus, the work done we by Mary is 2.215 J

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