On this question about work where an object is pulled up on an incline and we ne

Question

On this question about work where an object is pulled up on an incline and we need to find theta, where does this cosine theta come from in the formula W=Fd * cosine (theta)? I thought work was equal to force times distance. How can they just add cos (theta) there? I do not undestand how the same "W" can equal Fd and equal Fdcos(theta). Thanks!

Chapter 7: Work and Kinetic En James S. Walker, Physics, 4h Edition 85. Picture the Problem: The block slides horizontally but the ug force is F 45.0N inclined at an angle. v = 2.60 m/s Strategy: Use the definition of work to find the angle between the force and displacement vectors. Then set the work done by thg force equal to the change in kinetic energy of the block. 1.50 m Thought W-Fd. Where does the cos(theta) Solution: 1. (a) Solve 0.0J 42.2 50.0 J equation 7-3 for 2. (b) Solve equation 7-7 for-m-0 so Insight: Only the component of the force along the direction of the motion does any work. The vertical component of 5.0 N) (1.50 m 2W-2(5001) (2.60 m/s). the force reduces the nomal force a little

Explanation / Answer

Work is defined as the dot product between the force and displacement. So the expression for the work done is

W=F.d= Fdcos(theta)

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