Physics, 10.11.2019 11:31 karlaperez482

a 5.0 kg block is pushed 2.0 m at a constant
velocity up a vertical wall by a constant force
applied at an angle of 29.0° with the horizon-
tal, as shown in the figure.
the acceleration of gravity is 9.81 m/s.
29°
2 m
5 kg
drawing not to scale.
if the coefficient of kinetic friction between
the block and the wall is 0.30, find
a) the work done by the force on the block.
answer in units of j.
b) the work done by gravity on the block.
answer in units of j.
c) the magnitude of the normal force between
the block and the wall.
answer in units of n.

Answers: 2

Show answers

Answers

Answer from: pnpampanoa

(a) The work done by the force on the block 385.74 J.

(b) The work done by gravity on the block is 98 J.

the block and the wall is 192.87 N.

Given data:

Mass of block is, $m=5.0 \;\rm kg$ .

Distance covered due to push is, $d= 2.0 \;\rm m$ .

Angle with respect to the horizontal is, $\theta = 29 ^\circ$ .

The gravitational acceleration is, $g=9.8 \;\rm m/s^{2}$ .

The coefficient of kinetic friction between block and wall is, $\mu=0.3$ .

(a)

The expression for the work done by the constant force is,

$W = Fdcos \theta$

Here, F is the magnitude of constant force.

Taking equilibrium of forces along the vertical direction as,

$F_{net}=0\\Fsin\theta-f_{f}-mg=0\\Fsin\theta-\mu Fcos\theta-mg=0\\F=\dfrac{mg}{sin\theta-\mu cos\theta}$

Solving as,

$F=\dfrac{5 \times 9.8}{sin29-(0.30 \times cos29)}\\F=220.52 \;\rm N$

Then work done is,

$W = Fdcos \theta\\W = 220.52 \times 2 \times cos29\\W=385.74 \;\rm J$

Thus, the work done by the constant force is 385.74 J.

(b)

The work done by the gravity is,

$W'= -mgd\\W'= - 5 \times 9.8 \times 2\\W'=-98 \;\rm J$

Negative sign shows that applied force is in the opposite direction of motion.

Thus, the work done by the gravity is -98 J.

(c)

The magnitude of normal force is the cosine component of the constant force applied on the block. Then,

$N=Fcos\theta\\N= 220.52 \times cos 29\\N=192.87 \;\rm N$

Thus, the magnitude of normal force is 192.87 N.

Learn more about the work done in an inclined plane here:

link

a 5.0 kg block is pushed 2.0 m at a constant velocity up a vertical wall by a constant force applied

Answer from: starswarsfan

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

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 , $F_{friction}=\mu_{k}N=\mu_{k}Fcos(29\°)$ 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 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

When the applied force and the direction of motion form an angle the expression to calculate the Work 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

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

Answer from: Quest

its c its is a mixture of gases

Answer from: Quest

Milliseconds. . . . . .

Another question on Physics

Physics, 21.06.2019 17:00

What is the function of the commutator in a dc motor? a. to spin the magnet b. to change direct current into alternating current c. to change alternating current into direct current d. to spin the wire coil

Answers: 1

Answer

Physics, 22.06.2019 07:20

Diamond will scratch quartz. true or false

Answers: 2

Answer

Physics, 22.06.2019 18:30

In which situation is the acceleration of a car negative? question 5 options: a)the velocity of a car was 75 km/hr over 4 hours. b)the velocity of a car reduced from 50 km/hr over one minute. c)the velocity of a car increased from 40 km.h to 75 km/h over 15 minutes. d)the velocity of a car was 45 km/hr at 2: 00pm and at 4: 00pm its velocity was 85 km/ht.

Answers: 2

Answer

Physics, 22.06.2019 20:50

Asquare 1m x 1m plate is shown below. all four of its edges are kept at 0 co. at t=0the temperature distribution is given by: y xtsins n75)=.a) using separation of variables determine the temperature distribution t(x,y,t)along the plate at any time t.b) discretize the heat conduction equation +=22222 txtusing a second-order centered difference scheme for the two second derivatives and a first-order forward euler scheme for the first derivative. using von neumann stability analysis, determine the stability criterion for the proposed discretization scheme in terms of , hand t. solve for the maximum allowable time step tmax

Answers: 1

Answer

You know the right answer?

a 5.0 kg block is pushed 2.0 m at a constant
velocity up a vertical wall by a constant force

Questions