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tamaranim1 [39]

3 years ago

8

A 20 kg box has an initial velocity of 2 m/s starting at the bottom of a 30-degree inclined plane. A person pushes on the box di

rectly up the frictionless inclined plane so that it travels up the inclined plane at a constant velocity of 2 m/s. Calculate the how much is done by the person after 5 seconds have past.

1 answer:

Mice21 [21]3 years ago

5 0

Answer:

Explanation:

The box is moving with constant velocity so acceleration of box is zero . That means net force on the box is zero .

The weight component acting on box parallel to incline plane

= mg sin 30⁰ = 20 x 9.8 x sin 30 = 98 N

This force is acting down the plane , hence to make the net force zero acting on box , force exerted by person will also be 98 N up the incline .

Force exerted by person = 98 N

distance travelled in 5 s

= velocity x time

= 2 x 5 = 10 m

Work done by person

= 98 x 10

= 980 J .

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A 0.86 kg rock is projected from the edge of the top of a building with an initial velocity of 8.65 m/s at an angle 46◦ above th

Georgia [21]

Answer:

Height of the building = 11.4 m

Explanation:

As we know that the stone is projected at an angle 46 degree with speed 8.65 m/s

so the two components of the speed is given as

$v_x = 8.65 cos46$

$v_x = 6 m/s$

vertical component of the speed is given as

$v_y = 8.65 sin46$

$v_y = 6.22 m/s$

now we know that the ball strike at horizontal distance of 13.7 m

so we will have

$x = v_x t$

$13.7 = 6 t$

$t = 2.28 s$

now we know that in vertical direction ball will move under uniform gravity so we can use kinematics

$y = v_y t + \frac{1}{2}at^2$

$y = 6.22(2.28) - \frac{1}{2}(9.81)(2.28^2)$

$y = -11.4 m$

Height of the building = 11.4 m

3 0

3 years ago

Read 2 more answers

(7) A 2500 lbm car moving at 25 mi/hr is accelerated at a constant rate of 15 ft/s2 up to a speed of 50 mi/hr. What is the force

balandron [24]

Answer:

The force is $F = 1164.6\ lbf$

The time is $\Delta t = 2.44 \ s$

Explanation:

From the question we are told that

The mass of the car is $m = 2500 \ lbm$

The initial velocity of the car is $u = 25 \ mi/hr$

The final velocity of the car is $v = 50 \ mi/hr$

The acceleration is $a = 15 ft/s^2 = \frac{15 * 3600^2}{ 5280} = 36818.2 \ mi/h^2$

Generally the acceleration is mathematically represented as

$a = \frac{v-u}{\Delta t}$

=> $36818.2 = \frac{50 - 25 }{ \Delta t}$

=> $t = 0.000679 \ hr$

converting to seconds

$\Delta t = 0.0000679 * 3600$

=> $\Delta t = 2.44 \ s$

Generally the force is mathematically represented as

$F = m * a$

=> $F = 2500 * 15$

=> $F = 37500 \ \frac{lbm * ft}{s^2}$

Now converting to foot-pound-second we have

$F = \frac{37500}{32.2}$

=> $F = 1164.6\ lbf$

7 0

3 years ago

A box of mass 50 kg is pushed hard enough to set it in motion across a flat surface. Then a 99-N pushing force is needed to keep

salantis [7]

The box is kept in motion at constant velocity by a force of F=99 N. Constant velocity means there is no acceleration, so the resultant of the forces acting on the box is zero. Apart from the force F pushing the box, there is only another force acting on it in the horizontal direction: the frictional force $F_f$ which acts in the opposite direction of the motion, so in the opposite direction of F.
Therefore, since the resultant of the two forces must be zero,
$F-F_f=0$
so
$F=F_f$

The frictional force can be rewritten as
$F_f = \mu m g$
where $m=50 kg$ , $g=9.81 m/s^2$ . Re-arranging, we can solve this equation to find $\mu$ , the coefficient of dynamic friction:
$\mu = \frac{F}{mg}= \frac{99 N}{(50 kg)(9.81 m/s^2)} =0.20$

4 0

3 years ago

A negatively charged object is placed within a positive electric field what happens to the object

Anvisha [2.4K]

The object becomes neutralized in charge. The positive charge in the field neutralizes the negative charge in the object.

8 0

4 years ago

Read 2 more answers

A boy on a 1.9 kg skateboard initially at rest tosses a(n) 7.8 kg jug of water in the forward direction. if the jug has a speed

Tresset [83]

For this case we first think that the skateboard and the child are one body.
We have then:
1 = jug
2 = skateboard + boy
By conservation of the linear amount of movement:
M1V1i + M2V2i = M1V1f + M2V2f
Initial rest:
v1i = v2i = 0
0 = M1V1f + M2V2f
Substituting values
0 = (7.8) (3.2) + (M2) (- 0.65)
0 = 24.96 + M2 (-0.65)
-24.96 = (-0.65) M2
M2 = (-24.96) / (- 0.65) = 38.4 kg
Then, the child's mass is:
M2 = Mskateboard + Mb
Clearing:
Mb = M2-Mskateboard
Mb = 38.4 - 1.9
Mb = 36.5 Kg
answer:
the boy's mass is 36.5 Kg

4 0

3 years ago