Ask question

Ask question

All categories

I am Lyosha [343]

3 years ago

8

A student pushes on a 20.0 kg box with a force of 50 N at an angle of 30° below the horizontal. The box accelerates at a rate of

0.5 m/s2. What is the value of the friction force on the box?
200 N
50 N
43 N
33 N

2 answers:

anygoal [31]3 years ago

8 0

F - F(friction) = m.a
50cos(30) - F(friction) = 20 x 0.5
F(frictions) = 50cos(30) - 10 = 33.3 N

ikadub [295]3 years ago

6 0

Answer:

<u>Option-(D):</u> The value of the frictional force will be around 33 N.

Explanation:

So, we have a force, F of 50 N which is applied by a boy on a box of mass around 20 kg with an angle of 30 degree. We have the acceleration,a of 0.5 m/sec².As we know that each actions get some opposition and when we apply some force,F to an object there is a minimum amount of opposing force,F(friction) to it. And to find out the value of the frictional force,F(friction) we have the following equation to consider.

<u>F(applied) - Friction= m.a,</u>

F×cos(angle)-friction(tot)=mass(m).acceleration(a),
Now, putting all the required values inside the equation:
50×cos(30)-frictional force(F)=(20).(0.5),
50×(0.866)-frictional force(F)=10,
43.3-frictional force(F)=10,
frictional force(F)=43.3-10,
<u>F=33.3 N,</u>⇒Answer.

You might be interested in

During the annual shuffleboard competition, Renee gives her puck an initial speed of 8.12 m/s. Once leaving her stick, the puck

shutvik [7]

<h2>Answer:2.65 seconds</h2>

Explanation:

Let $a$ be the acceleration.

Let $u$ be the initial velocity.

Let $v$ be the final velocity.

Let $t$ be the time taken.

As we know from the equations of motion,

$v=u+at$

Given, $v=0\\u=8.12ms^{-1}\\a=-3.06ms^{-2}$

$0=8.12-3.06t$

$t=2.65sec$

6 0

3 years ago

A spring gun is made by compressing a spring in a tube and then latching the spring at

inn [45]

Answer:

a) $v=13.2171\,m.s^{-1}$

b) $H=8.9605\,m$

Explanation:

Given:

mass of bullet, $m=4.97\times 10^{-3}\,kg$

compression of the spring, $\Delta x=0.0476\,m$

force required for the given compression, $F=9.12 \,N$

(a)

We know

$F=m.a$

where:

a= acceleration

$9.12=4.97\times 10^{-3}\times a$

$a\approx 1835\,m.s^{-2}\\$

we have:

initial velocity, $u=0\,m.s^{-1}$

Using the eq. of motion:

$v^2=u^2+2a.\Delta x$

where:

v= final velocity after the separation of spring with the bullet.

$v^2= 0^2+2\times 1835\times 0.0476$

$v=13.2171\,m.s^{-1}$

(b)

Now, in vertical direction we take the above velocity as the initial velocity "u"

so,

$u=13.2171\,m.s^{-1}$

∵At maximum height the final velocity will be zero

$v=0\,m.s^{-1}$

Using the equation of motion:

$v^2=u^2-2g.h$

where:

h= height

g= acceleration due to gravity

$0^2=13.2171^2-2\times 9.8\times h$

$h=8.9129\,m$

is the height from the release position of the spring.

So, the height from the latched position be:

$H=h+\Delta x$

$H=8.9129+0.0476$

$H=8.9605\,m$

4 0

3 years ago

particle of mass 59 g and charge 51 µC is released from rest when it is 32 cm from a second particle of charge −14 µC. Determine

AleksAgata [21]

Answer:

The initial acceleration of the 59g particle is $1062.7\frac{m}{s^{2}}$

Explanation:

Newton's second laws relates acceleration (a), net force(F) and mass (m) in the next way:

$F=ma$ (1)

We already know the mass of the particle so we should find the electric force on it to use on (1), the magnitude of the electric force between two charged objects by Columb's law is:

$F=k\frac{\mid q_{1}q_{2}\mid}{r^{2}}$

with q1 and q2 the charge of the particles, r the distance between them and k the constant $k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}}$ . So:

$F=(9.0\times10^{9})\frac{\mid (51\times10^{-6})(-14\times10^{-6})\mid}{0.32^{2}}$

$F=62.7 N$

Using that value on (1) and solving for a

$a=\frac{F}{m}=\frac{62.7}{0.059}=1062.7\frac{m}{s^{2}}$

4 0

3 years ago

What terms are needed to completely describe velocity?

kotegsom [21]

In order to completely describe a velocity,
you need a speed and a direction.

4 0

3 years ago

What is the mass of an object that is accelerated at 25 m/s2 by a force of 135 N?

yKpoI14uk [10]

Answer:

<h3>The answer is 5.4 kg</h3>

Explanation:

The mass of the object can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question we have

$m = \frac{135}{25} = \frac{27}{5} \\$

We have the final answer as

<h3>5.4 kg</h3>

Hope this helps you

4 0

3 years ago