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Lapatulllka [165]

3 years ago

6

A cart loaded with bricks has a total mass of 9.13 kg and is pulled at constant speed by a rope. The rope is inclined at 24.7 ◦

degrees above the horizontal and the cart moves 15.1 m on a horizontal floor. The coefficient of kinetic friction between ground and cart is 0.597. The acceleration of gravity is 9.8 m/s 2 . How much work is done on the cart by the rope? Answer in units of kJ

1 answer:

blagie [28]3 years ago

6 0

Answer:

W = 0.63 KJ

Explanation:

Work (W) is defined as the point product of force (F) by the distance (d)the body travels due to this force.

W= F*d *cosα Formula (1)

F : force (N)

d : displacement (m)

α : angle between force and displacement

Newton's second law:

∑F = m*a Formula (2)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

We define the x-axis in the direction parallel to the movement of the cart on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the cart

W: Weight of the cart : In vertical direction

FN : Normal force : perpendicular to the floor

f : Friction force: parallel to the floor

T : tension Force, inclined at θ=24.7° above the horizontal

Calculated of the W

W= m*g

W= 9.13 kg* 9.8 m/s² = 89.47 N

x-y components o the tension force (T)

Tx = Tcosθ = T*cos 24.7° (N)

Ty = Tsin θ = T*sin 24.7° (N)

Calculated of the FN

We apply the formula (2)

∑Fy = m*ay ay = 0

FN +Ty- W = 0

FN = W-Ty

FN = 89.47-T*sin 24.7°

Calculated of the friction force (f)

f = μk*FN

f =(0.597)*( 89.47-T*sin 24.7° )

f= 53.41-0.249T

Calculated of the tension force of the rope (f)

We apply the formula (2) :

∑Fx = m*ax , ax= 0 ,because the speed of the cart is constant

Tx - f = 0

T*cos 24.7°-( 53.41 - 0.249T )= 0

T*cos 24.7° + 0.249T = 53.41

(1.1575)T = 53.41

T= (53.41) / (1.1575)

T= 46.14 N

Work done on the cart by the rope

We apply the formula (1)

W=T*d *cosα

W= (46.14 N)*(15.1 m) *(cos24.7)

W = 632.97 (N*m) = 632.97 (J)

W = 0.63 KJ

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Answer:

(a) the high of a hill that car can coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h is 47.6 m

(b) thermal energy was generated by friction is 1.88 x $10^{5}$ J

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Explanation:

given information:

m = 750 kg

initial velocity, $v_{0}$ = 110 km/h = 110 x 1000/3600 = 30.6 m/s $\frac{30.6^{2} }{2x9.8}$

initial height, $h_{0}$ = 22 m

slope, θ = 2.5°

(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h?

according to conservation-energy

EP = EK

mgh = $\frac{1}{2} mv_{0} ^{2}$

gh = $\frac{1}{2} v_{0} ^{2}$

h = $\frac{v_{0} ^{2} }{2g}$

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(b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction?

thermal energy = mgΔh

= mg (h - $h_{0}$ )

= 750 x 9.8 x (47.6 - 22)

= 188160 Joule

= 1.88 x $10^{5}$ J

(c) What is the average force of friction if the hill has a slope 2.5º above the horizontal?

f d = mgΔh

f = mgΔh / d,

where h = d sin θ, d = h/sinθ

therefore

f = (mgΔh) / (h/sinθ)

= 1.88 x $10^{5}$ /(22/sin 2.5°)

= 373 N

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Consider four point charges arranged in a square with sides of length L. Three of the point charges have charge q and one of the

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Answer: $F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]$

Explanation:

Given

Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge

Force due to the charge placed at diagonally opposite end on -q charge

$F_1=\frac{kq(-q)}{(L\sqrt{2})^2}$

where $L\sqrt{2}=$ Distance between the two charges

$F_1=-\frac{kq^2}{2L^2}$

negative sign indicates that it is an attraction force

Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge

$F_2=\frac{kq(-q)}{(L)^2}$

The magnitude of force by both the charge is same but at an angle of $90^{\circ}$

thus combination of two forces at 2 and 3 will be

$F'=\sqrt{2}\frac{kq^2}{2L^2}$

Now it will add with force due to 1 charge

Thus net force will be

$F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]$

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A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t

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a) $\omega=\frac{-mvR}{I+MR^2}$

b) $v=\frac{-mvR^2}{I+MR^2}$

Explanation:

a)

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At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

$L_1=0$

Later, after the girl throws the rock, the angular momentum will be:

$L_2=(I_M+I_g)\omega +L_r$

where:

$I$ is the moment of inertia of the merry-go-round

$I_g=MR^2$ is the moment of inertia of the girl, where

M is the mass of the girl

R is the distance of the girl from the axis of rotation

$\omega$ is the angular speed of the merry-go-round and the girl

$L_r=mvR$ is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

$L_1=L_2$

So we find:

$0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}$

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

b)

The linear speed of a body in rotational motion is given by

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

$\omega=\frac{-mvR}{I+MR^2}$ is the angular speed

$r=R$ is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

$v=\omega R=\frac{-mvR^2}{I+MR^2}$

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