Question

11 of 12

Constants | Periodic Table

Part A

How much work is done by the horizontal force FP = 150 N on the 18-kg block of the figure(Figure 1) when the force pushes the block 5.0 m up along the 32 ? frictionless incline?

Express your answer using two significant figures.

Wp =

  J  

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Part B

How much work is done by the gravitational force on the block during this displacement?

Express your answer using two significant figures.

WG =

  J  

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Part C

How much work is done by the normal force?

Express your answer using two significant figures.

WN =

  J  

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Part D

What is the speed of the block (assume that it is zero initially) after this d is placement? [Hint:Work-energy involves net work done.]

Express your answer using two significant figures.

v =

  m/s  

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Answer 1

Here ,

part A)

work done by the horizontal force , FP = FP * d * cos(theta)

work done by the horizontal force , FP = 150 * 5 * cos(32 degree)

work done by the horizontal force , FP = 636 J

part B)

work done by gravitational force = - m * g * d * sin(theta)

work done by gravitational force = -18 * 9.8 * 5 * sin(32 degree)

work done by gravitational force = -467 J

part C)

as force is perpendicular to displacement to incline

work done by normal force = 0 J

part D)

let the speed of the block is v

Using work energy theorem

change in kinetic energy = total work done

636 - 467 = 0.50 * 18 * v^2

solving for v

v = 4.33 m/s

the speed of the block is 4.33 m/s