1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
lord [1]
3 years ago
11

A 5 kg wooden block sitson a flat straight-away12 meters fromthe bottom of an infinitely long ramp, which has an angle of 20 deg

rees. The straightaway has a coefficient of kinetic friction of 0.4and the ramp of 0.25. While the block is on the flat section, you exert a 100 N forceat an angle of 45 degrees above the horizontal, propelling the block towards the ramp.
a.What is the velocity of the block right before it begins going up the ramp?

b.How far up the ramp (the distance along the hypotenuse) does the block travel before stopping?
Physics
1 answer:
saveliy_v [14]3 years ago
3 0

Answer:

(a) 19.71801m/s Velocity just before going up the ramp.

(b) 74.56338m.

Explanation:

We will solve it in two parts, first we will calculate time that 5kg wooden block would take to just reach ramp and with this time we will calculate final velocity that the wooden block would have in this time.

Second, we will calculate the component of velocity vector along inclined plane and the time that it would take for velocity to be 0 meters/s then with this time we will calculate the distance that inclined plane would travel along inclined plane.

Following formulas will be used.

                                  x(t) = \frac{1}{2} t^2 = 12m =16.2m/s^2 t^2

                                 F =ma

                                 V(t) = V_{o} +at

                                 x(t) = x_{0} +v_{0}t+\frac{1}{2}a t^2

(a) Calculating velocity right before going up the ramp.

 Wooden block is going on a straightaway and has net for on it.

         F_{n} =F-F_{s} = F-uF_{n}  = 100N-0.4*9.8m/s^2*5kg =81N

     and this force produces acceleration of

      a = \frac{F}{m}=\frac{81}{5} =16.2m/s^2 .

With this acceleration, wooden block would reach at the foot of ramp in.

          x(t) = 12m = 16.2m/s^2*t^2

         t = 1.217s

and final velocity will be

v(t) = v_{0}+at = 0+16.2m/s^2*1.2171s = 19.7180m/s.

this velocity of wooden box just before going up the ramp.

(b) How far up the ramp will the wooden block go before stopping.

Ramp is at 20° relative to horizontal therefore velocity along the ramp that the wooden block would have will be.

                              V= V_{h}cos(20) = 18.5288m/s

and deceleration along the ramp is

                              a = \frac{F_{s} }{m}

 Where F_{s} force of friction along the inclined plane.

F_s =  uF_n = u*m*a

a = 9.8m/s^2*cos(20) = 9.2089m/s^2

is a component of g along normal of the inclined plane.

                               F_{s} = 0.25*5kg*9.2089m/s^2

                              = 11.5112N

                              a = \frac{11.5112N}{5kg} = 2.3022m/s^2

And with this deceleration time needed to get wooded block to stop is.

                     v(t) = v_o-at = 18.5288m/s-2.3022m/s^2*t = 0

                        t = \frac{18.5288m/s}{2.3022m/s^2} =8.04813s

 and in that time wooden block would travel

   x(8.04813s) = 18.52881m/s *8.04813s-\frac{1}{} 2.3022m/s^2*(8.0481)^2=74.56338m

This is how up wooden box will go before coming to stop.

You might be interested in
Moving the probe 1 cm towards the non-grounded electrode changes the value the potential from about 0.90 V to about 1.2 V. Expla
svp [43]

Answer:

-30 N/C

Explanation:

Since the potential changes from 0.90 V to 1.2 V when I move the probe 1 cm closer to the non-grounded electrode, the electric field is the gradient between the two points is given by E = -ΔV/Δx where ΔV = change in electric potential and Δx = distance of potential change = 1 cm = 0.01 m

Now ΔV = final potential - initial potential = 1.2 V - 0.90 V = 0.30 V

Since E = -ΔV/Δx

substituting the values of the variables into the equation, we have

E = -ΔV/Δx

E = -0.30 V/0.01 m

E = -30 V/m

Since 1 V/m = 1 N/C.

E = -30 N/C

So, the average electric field is -30 N/C

6 0
3 years ago
What force is necessary to accelerate a 2500 kg care from rest to 20 m/s over 10 seconds?
EleoNora [17]
Force = mass x acceleration
force = 2500kg x (20m/s / 10m/s)
force = 2500kg x 2m/s^2
force = 5000kg m/s^2 = 5kN

i hope this is right (^^)
4 0
2 years ago
the very high voltage needed to create a spark across the spark plug is produced at the a. transformer's primary winding. b. tra
Karo-lina-s [1.5K]
I think the correct answer from the choices listed above is option B. The very high voltage needed to create a spark across the spark plug is produced at the  transformer's secondary winding. <span>The secondary coil is engulfed by a powerful and changing magnetic field. This field induces a current in the coils -- a very high-voltage current.</span>
5 0
3 years ago
Read 2 more answers
Is the relationship between velocity and centripetal force a direct, linear or nonlinear square relationship?
Svet_ta [14]

Answer:

non linear square relationship

Explanation:

formula for centripetal force is given as

a = mv^2/r

here a ic centripetal acceleration , m is mass of body moving in circle of radius r and v is velocity of body . If m ,and r are constant we have

a = constant × v^2

a α v^2

hence non linear square relationship

5 0
3 years ago
A rescue plane flying horizontally at 72.6 m / s spots a survivor in the ocean 182 m directly below and releases an emergency ki
Mila [183]

Answer:

547 m

Explanation:

From law of motion

s = ut + ½at²

Where "t" is Time taken to reach Earth

s= distance= 182 m

a= vertical acceleration = 5.82 m / s 2

U= initial velocity in vertical position = 0

182= ½ × 5.82t²

t²=( 2× 182)/ 5.82

= 364/5.82

= 62.54

t= √62.54

t= 7.908s

horizontal distance travelled = speed x time

Horizontal speed= 72.6 m / s

horizontal distance travelled =72.6× 7.908

= 547 m

Hence, the survivor will it hit the waves at 547 m away

3 0
2 years ago
Other questions:
  • What effect does dropping the sandbag out of the cart at the equilibrium position have on the amplitude of your oscillation? Vie
    9·2 answers
  • Which statement best describes the explanatory powers of hypotheses and theories?
    6·2 answers
  • Which type of reaction has the general formula of AB + CD → AD + CB?
    12·1 answer
  • A constant voltage of 12.00 V has been observed over a certain time interval across a 1.20 H inductor. The current through the i
    5·1 answer
  • Floodplains are most often found for rivers that exist on
    9·2 answers
  • Which one of the following statements is not a characteristic of a plane mirror?
    6·1 answer
  • A block of mass 2 kg is placed on the floor. The coefficient of static friction is 0.4. A horizontal force of 2.5 N is applied o
    15·1 answer
  • He wrote a book of aerobics
    10·2 answers
  • Help plsssss<br><br><br><br><br> ..........
    14·1 answer
  • Abdou was explaining to a classmate that graphite is a good lubricant because it is bonded in layers that easily slip over each
    14·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!