Question

the component of $vec{f}$ parallel to $deltavec{x}$ points in the same direction as $deltavec{x}$, so that the work is positive, and so that we have
$w_{app}=fdelta xcos\theta = fdcos\theta$
$=(16 n)(6.65 m)cos35^{circ}=87.2 j$,
where $w_{app}$ is the work done by the applied force $vec{f}$.
step 2
(b) the normal force is perpendicular to the displacement, so that the work done by the normal force is
your response differs significantly from the correct answer. rework your solution from the beginning and check each step carefully. j,
(c) the downward force $mvec{g}$ due to the weight of the block is also perpendicular to the displacement so that the work done by this force is
your response differs significantly from the correct answer. rework your solution from the beginning and check each step carefully. j. enter an exact number
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Explanation:

Step1: Recall work - formula

The work done by a force $\vec{F}$ over a displacement $\vec{d}$ is given by $W = Fd\cos\theta$, where $\theta$ is the angle between the force and the displacement.

Step2: Analyze normal - force work

The normal force $\vec{N}$ is perpendicular to the displacement $\vec{d}$, so $\theta = 90^{\circ}$. Then $W_N=F_Nd\cos90^{\circ}$. Since $\cos90^{\circ}=0$, $W_N = 0$.

Step3: Analyze weight - force work

The weight force $mg$ is perpendicular to the horizontal displacement, so $\theta = 90^{\circ}$. Then $W_{mg}=mgd\cos90^{\circ}$. Since $\cos90^{\circ}=0$, $W_{mg}=0$.

Answer:

(b) 0 J
(c) 0 J