Net Work & Kinetic Energy Video Lessons

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Problem: A. How much work is done by the horizontal force FP = 150 N on the 18-kg block of the figure when the force pushes the block 6.0 m up along the 32 frictionless incline?B. How much work is done by the gravitational force on the block during this displacement?C. How much work is done by the normal force?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.]

FREE Expert Solution

Work done by the force in pushing an object over a distance d is given by:

W=F·d=Fdcosθ

A.

In this case, the applied force, Fp = 150N

Displacement, d = 6.0 m

θ = 32°

Substituting to our work equation gives:

W=Fdcosθ=(150)(6.0)cos(32°)

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Problem Details

A. How much work is done by the horizontal force FP = 150 N on the 18-kg block of the figure when the force pushes the block 6.0 m up along the 32 frictionless incline?

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

C. How much work is done by the normal force?

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.]

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