Question

A 5.0 kg block is pushed 2.0 m at a constant

Answer 1

Answers:

a) 385.74 J

b) -98.1 J

c) 192.87 N

Explanation:

A figure is attached in order to understand better the question. In addition, we have the following data:

$m=5 kg$ is the mass of the block

$d=2 m$ is the distance the block is pushed at constant velocity

$\theta=29\°$ is the angle at wich the force $F$ is applied

$g=9.81 m/s^{2}$ is the acceleration due gravity

$\mu_{k}=0.3$ is the coefficient of kinetic friction between  the block and the wall

Now, if we draw a free body diagram of this situation we will have the following:

$\sum{F_{netx}}=-N+F_{x}=0$

Where $N$ is the Normal force and $F_{x}=Fcos(29\°)$ is the horizontal component of the applied force $F$

Then: $N=Fcos(29\°)$ (1)

$\sum{F_{nety}}=F_{y}-F_{friction}-Fg=0$

Where $F_{y}=Fsin(29\°)$ is the vertical component of the applied force $F$, $F_{friction}=\mu_{k}N=\mu_{k}Fcos(29\°)$ is the Friction force and $Fg=mg$ is the force due gravity (the weight of the block)

Then: $Fsin(29\°)-\mu_{k}Fcos(29\°)-mg=0$ (2)

Finding $F$ from (2):

$F=\frac{mg}{sin(29\°)-\mu_{k}cos(29\°)}$ (3)

$F=\frac{(5 kg)(9.81 m/s^{2})}{sin(29\°)-0.3cos(29\°)}$ (4)

$F=220.52 N$ (5)

a) Now we can calculate the work done by the force $F$:

When the applied force and the direction of motion form an angle the expression to calculate the Work $W$ is:  

$W=Fdcos{\theta}$ (6)

$W=(220.52 N)(2 m)cos{29\°}$ (7)

$W=385.74 J$ (8) This is the work done by $F$

b) The work done by gravity on the block

In this case the work equation is:

$Wg=Fg d cos{\alpha}$ (9)

Where:

$Fg=mg=(5 kg)(9.81 m/s^{2})=49.05 N$ is the gravity force on the block

$\alpha=180\°$ is the angle between the gravity force and the direction of motion

$Wg=(49.05 N)(2 m) cos(180\°)$ (10)

$Wg=-98.1 J$ (11) Note the work is negative because the applied force is in the opposite direction of motion

c) The magnitude of the normal force

From (1) we know the Normal force is:

$N=Fcos(29\°)$ (1)

Since $F=220.52 N$, we have:

$N=220.52 N cos(29\°)$ (12)

$N=192.87 N$ (13) This is the Normal force