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3 years ago

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A dog is pulling a sled by applying a force of 75 newtons on it. The angle of force from the ground is 45°. If the sled moves 15

meters, how much work is done by the dog on the sled?

1 answer:

Kitty [74]3 years ago

4 0

Work done = force x distance (where the force is in the direction of the distance travelled)
We must resolve the force so that it is horizontal to the ground
force = 75cos45
Then simply plug into the formula, so:
work done = 75cos45 x 15 = 795.49... Nm = 795 Nm (3 sig. fig.)

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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

saveliy_v [14]

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.

3 0

3 years ago

PLEASE HELP! I don't get it at all! Speed is one thing; distance is another. Where is the arrow you shoot up at 50m/s when it ru

LuckyWell [14K]

I got you b, V(final)^2=V(initial+2acceleration*displacement
So this turns to (0m/s)^2=(50m/s)^2+2(9.8)(d) so just flip it all around to isolate d so you get
-(50m/s)^2/2(9.8) = d so you get roughly 12.7555 meters up

4 0

3 years ago

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The net horizontal force on a car is 981 N. The car has a mass of 1550 kg and the force is applied when the car has a speed of 2

viktelen [127]

Answer:

Distance, d = 778.05 m

Explanation:

Given that,

Force acting on the car, F = 981 N

Mass of the car, m = 1550 kg

Initial speed of the car, v = 25 mi/h = 11.17 m/s

We need to find the distance covered by car if the force continues to be applied to the car. Firstly, lets find the acceleration of the car:

$F=ma\\\\a=\dfrac{F}{m}\\\\a=\dfrac{981}{1550}\\\\a=0.632\ m/s^2$

Let d is the distance covered by car. Using second equation of motion as :

$d=ut+\dfrac{1}{2}at^2\\\\d=11.17\times 35+\dfrac{1}{2}\times 0.632\times (35)^2\\\\d=778.05\ m$

So, the car will cover a distance of 778.05 meters.

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3 years ago

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Harlamova29_29 [7]

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3 years ago

A solid sphere, a solid disk, and a thin hoop are all released from rest at the top of the incline (h0 = 20.0 cm).

Ede4ka [16]

Answer:

a. The object with the smallest rotational inertia, the thin hoop

b. The object with the smallest rotational inertia, the thin hoop

c. The rotational speed of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

Explanation:

a. Without doing any calculations, decide which object would be spinning the fastest when it gets to the bottom. Explain.

Since the thin has the smallest rotational inertia. This is because, since kinetic energy of a rotating object K = 1/2Iω² where I = rotational inertia and ω = angular speed.

ω = √2K/I

ω ∝ 1/√I

since their kinetic energy is the same, so, the thin hoop which has the smallest rotational inertia spins fastest at the bottom.

b. Again, without doing any calculations, decide which object would get to the bottom first.

Since the acceleration of a rolling object a = gsinФ/(1 + I/MR²), and all three objects have the same kinetic energy, the object with the smallest rotational inertia has the largest acceleration.

This is because a ∝ 1/(1 + I/MR²) and the object with the smallest rotational inertia has the smallest ratio for I/MR² and conversely small 1 + I/MR² and thus largest acceleration.

So, the object with the smallest rotational inertia gets to the bottom first.

c. Assuming all objects are rolling without slipping, have a mass of 2.00 kg and a radius of 3.00 cm, find the rotational and translational speed at the bottom of the incline of any one of these three objects.

We know the kinetic energy of a rolling object K = 1/2Iω² + 1/2mv² where I = rotational inertia and ω = angular speed, m = mass and v = velocity of center of mass = rω where r = radius of object

The kinetic energy K = potential energy lost = mgh where h = 20.0 cm = 0.20 m and g = acceleration due to gravity = 9.8 m/s²

So, mgh = 1/2Iω² + 1/2mv² = 1/2Iω² + 1/2mr²ω²

Let I = moment of inertia of sphere = 2mr²/5 where r = radius of sphere = 3.00 cm = 0.03 m and m = mass of sphere = 2.00 kg

So, mgh = 1/2Iω² + 1/2mr²ω²

mgh = 1/2(2mr²/5 )ω² + 1/2mr²ω²

mgh = mr²ω²/5 + 1/2mr²ω²

mgh = 7mr²ω²/10

gh = 7r²ω²/10

ω² = 10gh/7r²

ω = √(10gh/7) ÷ r

substituting the values of the variables, we have

ω = √(10 × 9.8 m/s² × 0.20 m/7) ÷ 0.03 m

= 1.673 m/s ÷ 0.03 m

= 55.77 rad/s

≅ 55.8 rad/s

So, its rotational speed is 55.8 rad/s

Its translational speed v = rω

= 0.03 m × 55.8 rad/s

= 1.67 m/s

So, its rotational speed is of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

6 0

2 years ago