Question

(b) how much work is done by the friction force and by $vec{f}$? (dont forget the signs. use the following as necessary: $f$ for the magnitude of $vec{f}$, $g$, $m$, $x$, $\theta$ and $mu_{k}$.)
$w_{f_{k}}=mu_{k}(mg - f-sin(f))$
$w_{f}=fxcos(f)$
(c) identify all the forces that do no work on the block. (select all that apply.)
$vec{mg}$
friction $vec{f}_{k}$
$vec{f}cos(\theta)$
$vec{f}sin(\theta)$
$vec{n}$
(d) let $m = 2.00$ kg, $x = 3.80$ m, $\theta = 4.2^{circ}$, $f = 14.5$ n, and $mu_{k}=0.400$, and find the answers to parts (a) and (b). (include appropriate signs. enter the friction force in n and work in j.)
$f_{k}=- 15.0$
the response you submitted has the wrong sign. n
$w_{f}=41.2$
your response differs from the correct answer by more than 10%. double check your calculations. j
$w_{f_{k}}=-15.0$
your response differs from the correct answer by more than 10%. double check your calculations. j

Explanation:

Step1: Recall work - formula

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

Step2: Analyze work done by friction force

The friction force $\vec{f}_k=\mu_k N$. The normal force $N$ needs to be determined from the force - balance in the direction perpendicular to the displacement. Assuming the block is moving on a horizontal surface and a force $\vec{F}$ is applied at an angle $\theta$ with the horizontal, $N = mg - F\sin\theta$. The friction force acts in the opposite direction of the displacement, so $W_{f_k}=-\mu_k(mg - F\sin\theta)x$.

Step3: Analyze work done by $\vec{F}$

The component of the force $\vec{F}$ in the direction of the displacement is $F\cos\theta$. So the work done by $\vec{F}$ is $W_F = Fx\cos\theta$.

Step4: Determine non - working forces

A force does no work if the angle between the force and the displacement is $90^{\circ}$ (since $\cos90^{\circ}=0$). The weight $\vec{mg}$ and the normal force $\vec{n}$ act perpendicular to the horizontal displacement of the block, so they do no work.

Step5: Calculate numerical values

Given $m = 2.00\ kg$, $x = 3.80\ m$, $\theta = 4.2^{\circ}$, $F = 14.5\ N$, $\mu_k=0.400$, $g = 9.8\ m/s^2$.
First, find the normal force $N=mg - F\sin\theta=2\times9.8-14.5\times\sin(4.2^{\circ})=19.6 - 14.5\times0.073=19.6 - 1.0685 = 18.5315\ N$.
The friction force $f_k=\mu_k N=0.4\times18.5315 = 7.4126\ N$. The work done by friction $W_{f_k}=- 7.4126\times3.8=-28.1679\ J$.
The work done by $\vec{F}$ is $W_F = Fx\cos\theta=14.5\times3.8\times\cos(4.2^{\circ})=14.5\times3.8\times0.9972=54.38\ J$.

Answer:

(b) $W_{f_k}=-\mu_k(mg - F\sin\theta)x$, $W_F = Fx\cos\theta$
(c) $\vec{mg}$, $\vec{n}$
(d) $f_k=-7.41\ N$, $W_F = 54.38\ J$, $W_{f_k}=-28.17\ J$