Question

PHY 220 Laboratory Experiment 9 Prelab: WORK AND ENERGY Name In the lab you will make use of a setup that is similar to the situation shown below. The picture shows a block of mass M on an incline elevated an angle 0above the horizontal. The block starts from rest at point P. Neglect friction between the block and the incline.) 1. Block starts from rest at P incline is frictioniess a. Draw a free body diagram for the block after it has begun to move down the incline. Acceleration of block In the box at the right, draw an arrow to indicate the direction of the acceleration of the glider. Explain why you chose this direction. b. c. In the box at the right, draw an arrow to indicate the direction of the net Net force on block force on the glider. Explain why you chose this direction. Write an expression for the magnitude of the net force on the block in terms of given or known quantities. (Hint: Think about the forces acting parallel to and perpendicular to the incline.) Explain. d. As the block travels from point P to point R, does the magnitude of the net force exerted on the block increase, decrease, or remain constant? Explain. e. 2. The work done on an object by a force F while the object travels a distance d is defined as 0) where 0 is the angle between the force and the distance traveled. continued on other side) P9-1

I got the first side done and included the answers, I just need the second part

Answer 1

Hi,

Hope you are doing well.

PART 2:

(a)

Refer the free body diagram provided below to understand the concept easier:

Refer the free body diagram provided below to understand the concept easier: From free body diagram it is clear that the normal force N cancels the force mg cos theta Therefore Net force acting on the body is, F_NET - mg sin theta From free body diagram, angle between the net force and the displacement is given by, theta-FD = 0 degree Displacement between the point P and R is given by, d = L Therefore Net Work done on the object is given by, W-NET = F_NET d cos theta_FD = (mg sin theta) middot L. cos theta degree W_Net = + mg L sin theta Therefore Network done on the body is positive.

FORCE	WORK ON BLOCK (Positive/ Negative/ Zero)	EXPLANATION
	Zero	Angle between displacement and force ( $\theta_{Fd}$ ) is 90o. ie; displacement vector and force vector are perpendicular. $\therefore cos\theta_{Fd}=\cos 90^{\circ}=0$ $\therefore W=Fd \cos\theta_{Fd}=Fd\cos 90^{\circ}=0$
$mg\cos\theta$	Zero	Angle between displacement and force ( $\theta_{Fd}$ ) is 90o. ie; displacement vector and force vector are perpendicular. $\therefore cos\theta_{Fd}=\cos 90^{\circ}=0$ $\therefore W=Fd \cos\theta_{Fd}=Fd\cos 90^{\circ}=0$
$mg\sin\theta$	Positive	Angle between displacement and force ( $\theta_{Fd}$ ) is 0o. ie; displacement vector and force vector are parallel(same direction). $\therefore cos\theta_{Fd}=\cos 0^{\circ}=1$ $\therefore W=Fd \cos\theta_{Fd}=Fd\cos 0^{\circ}=+Fd$

(b)

From free body diagram it is clear that the normal force cancels the force $mg\cos\theta$ .

$\therefore$ Net force acting on the body is,

$F_{NET}=mg\sin\theta$

From free body diagram, angle between the net force and the displacement is given by,

$\theta_{Fd}=0^{\circ}$

Displacement between the point P and R is given by,

$d=L$

$\therefore$ Net Work done on the object is given by,

$W_{NET}=F_{NET}d\cos\theta_{Fd}=(mg\sin\theta).L.\cos 0^{\circ}$

$W_{NET}=+mgL \sin\theta$

$\therefore$ Net work done on the body is positive.

(Note: In this case the sign of the work done only depends on the angle between the net force and the displacement. So if cosine of angle between them is positive then the net work done is positive or else negative. If angle between them is 90^o, then the work done will be ZERO.)

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