PHYSICS COLLEGE

An object is being acted upon by three forces and moves with a constant velocity. One force is 60.0 N along the x-axis, the second in 75.0 N along the y-axis. What is the magnitude of the third force?

Answers

Answer 1

Answer:

Answer:

96.05 N

Explanation:

From Vector,

The two forces acting along the x and y axis are perpendicular,

Fr = √(60²+75²) .............. Equation 1

Where Fr is the result of the two forces

Fr = √(3600+5625)

Fr = √(9225)

Fr = 96.05 N.

Note: Since the object moves with a constant velocity when it is acted upon by the three forces, The acceleration is zero and as such the resultant of the forces is equal to zero.

Therefore,

Ft = Fr+F3................... Equation 2

Where Ft = Total resistance of the three forces, F3 = magnitude of the third force.

make F3 the subject of the equation,

F3 = Ft-Fr

Given: Ft = 0 N, Fr = 96.05 N.

Substitute into equation 2

F3 = 0-96.05

F3 = -96.05 N.

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Answers

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While a roofer is working on a roof that slants at 37.0 ∘∘ above the horizontal, he accidentally nudges his 92.0 NN toolbox, causing it to start sliding downward, starting from rest. If it starts 4.25 m from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is 22.0 N?

Answers

Answer:

The speed is $v =8.17 m/s$

Explanation:

From the question we are told that

The angle of slant is $\theta = 37.0^o$

The weight of the toolbox is $W_t = 92.0N$

The mass of the toolbox is $m = (92)/(9.8) = 9.286kg$

The start point is $d = 4.25m$ from lower edge of roof

The kinetic frictional force is $F_f = 22.0N$

Generally the net work done on this tool box can be mathematically represented as

$Net \ work done = Workdone \ due \ to \ Weight + Workdone \ due \ to \ Friction$

The workdone due to weigh is = $mgsin \theta * d$

The workdone due to friction is = $F_f \ cos\theta * d$

Substituting this into the equation for net workdone

$W_(net) = mgsin\theta * d + F_f \ cos \theta *d$

Substituting values

$W_(net) = 92 * sin (37) * 4.25 + 22 cos (37) * 4.25$

$= 309.98 J$

According to work energy theorem

$W_(net) = \Delta Kinetic \ Energy$

$W_(net) = (1)/(2) m (v - u)^2$

From the question we are told that it started from rest so u = 0 m/s

$W_(net) = (1)/(2) * m v^2$

Making v the subject

$v = \sqrt{(2 W_(net))/(m) }$

Substituting value

$v = \sqrt{(2 * 309.98)/(9.286) }$

$v =8.17 m/s$

Circuit A in a house has a voltage of 218 V and is limited by a 45-A circuit breaker. Circuit B is at 120.0 V and has a 25-A circuit breaker.What is the ratio of the maximum power delivered by circuit A to that delivered by circuit B?

Answers

Answer:

3.27

Explanation:

Electric Power: This can be defined as the rate at which electric energy is consumed. The unit of power is Watt (W).

Mathematically, electric power is represented as

P = VI ..................................... Equation 1.

Where P = power, V = voltage, I = Current.

For Circuit A,

P₁ = V₁I₁ ................................... Equation 2

Where P₁ = maximum power delivered by circuit A, V₁ = Voltage of circuit A, I₁ = circuit breaker rating of circuit A.

Given: V₁ = 218 V, I₁ = 45 A.

Substituting into equation 2

P₁ = 218×45

P₁ = 9810 W.

For Circuit B,

P₂ = V₂I₂............................. Equation 3

Where P₂ = maximum power delivered by the circuit B, V₂ = voltage of circuit B, I₂ = circuit breaker rating of circuit B

Given: V₂ = 120 V, I₂ = 25 A.

Substitute into equation 3

P₂ = 120(25)

P₂ = 3000 W.

Ratio of maximum power delivered by circuit A to that delivered by circuit B = 9810/3000

= 3.27.

Thus the ratio of maximum power delivered by circuit A to circuit B = 3.27

Susan’s 10 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30° above the floor. The tension is a constant 30 N and the coefficient of friction is 0.20. Use work and energy to find Paul’s speed after being pulled 3.0 m.

Answers

Final answer:

In this problem, we have calculated the work done by Susan pulling her baby brother on a mat and the work done against friction. The net work done, which is the work done by Susan's pulling minus the work done against friction, is transformed into kinetic energy, giving us the baby's speed after being pulled 3m, which is approximately 1.95 m/s.

Explanation:

To answer this question, we first need to calculate the work done by Susan when she pulls the mat over the distance of 3.0 meters. The angle at which the rope is pulled does make a difference in this calculation. The force that is actually contributing to the work is the horizontal component of the tension, which can be determined by the equation Fh = F cos θ which equals 30N * cos30 = 25.98N.

The work done, W, is equal to this force multiplied by the displacement, so W = Fd = 25.98N * 3m = 77.94 Joules.

Next, we need to calculate the work done against friction. The force of friction is calculated as Ff = µN. Here N is the normal force, which is equal to the weight of the baby, so N = mg = 10kg * 9.8m/s² = 98N. The force of friction then is Ff = µN = 0.20 * 98N = 19.6N. The work done against friction is Wf = Ff * d = 19.6N * 3m = 58.8 Joules.

The net work done on the baby is the work done by Susan minus the work done against friction, so Wnet = W - Wf = 77.94J - 58.8J = 19.14 Joules. This net work is equal to the change in kinetic energy of the baby, ∆K, since Kinitial = 0 (Paul starts at rest), the work done is all transformed into final kinetic energy. So ∆K = 19.14J.

The kinetic energy of an object is given by the equation K = 1/2 mv², so we have 19.14J = 1/2 * 10kg * v². Solving for v gives us roughly v = 1.95 m/s. Therefore, the speed of the baby after being pulled 3 meters is approximately 1.95 m/s.

Learn more about Physics: Work and Energy here:

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Final answer:

To determine Paul's speed, we must calculate the net work done on him using the work-energy theorem. This includes the work done by Susan and the work done against friction. Paul’s speed after being pulled 3.0 m is approximately 1.96 m/s.

Explanation:

Solving this problem involves understanding the work-energy theorem and forces. First, let's calculate the work done. The work done by the force Susan applies (W1) is the product of the tension (T), the distance (d), and the cosine of the angle (θ). W1 = T * d * cos(θ) = 30N * 3.0m * cos(30) = 77.94J.

Next, the work done against friction (W2) is the product of the frictional force and the distance, which is µmgd. Here, µ is the coefficient of friction (0.20), m (10kg) is the mass of the baby, g (9.8m/s2) is the acceleration due to gravity, and d is the distance (3.0 m). W2 = µmgd = 0.20 * 10kg * 9.8m/s2 * 3.0m = 58.8J.

According to the work-energy theorem, the net work done on an object is equal to the change in its kinetic energy. Therefore, the final kinetic energy (and thus the final speed) of Paul will be the initial kinetic energy plus the net work done on him. His initial speed is assumed to be zero, hence the initial kinetic energy is zero. The net work done on him is W = W1 - W2= 77.94J - 58.8J = 19.14J. Setting this equal to the final kinetic energy, (1/2)mv2, allows us to solve for the final speed, v = sqrt((2 * W)/m) = sqrt((2 * 19.14J)/10kg) = 1.96 m/s approximately.

Learn more about the Work-Energy Theorem here:

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A 220 kg crate hangs from the end of a rope of length L = 14.0 m. You push horizontally on the crate with a varying force F to move it 4.00 m to the side.(a) What is the magnitude of F when the crate is in this final position? During the crate's displacement, what are
(b) the total work done on it,
(c) the work done by the gravitational force on the crate, and
(d) the work done by the pull on the crate from the rope?
(e) Knowing that the crate is motionless before and after its displacement, use the answers to (b), (c), and (d) to find the work your force F does on the crate.
(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)?

Answers

Answer:

(a) magnitude of F = 797 N

(b)the total work done W = 0

(c)work done by the gravitational force = -1.55 kJ

(d)the work done by the pull = 0

(e) work your force F does on the crate = 1.55 kJ

Explanation:

Given:

Mass of the crate, m = 220 kg

Length of the rope, L = 14.0m

Distance, d = 4.00m

(a) What is the magnitude of F when the crate is in this final position

Let us first determine vertical angle as follows

=> $Sin \theta = (d )/(L)$

=> $\theta = Sin^(-1) (d)/(L)$ =

Now substituting thje values

=> $\theta = Sin^(-1) (4)/(12)$ =

=> $\theta = Sin^(-1) (1)/(3)$

=> $\theta = Sin^(-1)(0.333)$

=> $\theta = 19.5^(\circ)$

Now the tension in the string resolve into components

The vertical component supports the weight

=> $Tcos\theta = mg$

=> $T = (mg)/(cos\theta)$

=> $T = (230 * 9.8 )/(cos(19.5))$

=> $T = (2254 )/(cos(19.5))$

=> $T = (2254 )/(0.9426)$

=>T =2391N

Therefore the horizontal force

$F = TSin(19.5)$

F = 797 N

b) The total work done on it

As there is no change in Kinetic energy

The total work done W = 0

c) The work done by the gravitational force on the crate

The work done by gravity

Wg = Fs.d = - mgh

Wg = - mgL ( 1 - Cosθ )

Substituting the values

= $-230 * 9.8* 12 ( 1 - cos(19.5) )$

= $-230 * 9.8* 12 ( 1 - 0.9426) )$

= $-230 * 9.8* 12 (0.0574)$

= $-230 * 9.8* 0.6888$

= $-230 * 6.750$

= -1552.55 J

The work done by gravity = -1.55 kJ

d) the work done by the pull on the crate from the rope

Since the pull is perpendicular to the direction of motion,

The work done = 0

e)Find the work your force F does on the crate.

Work done by the Force on the crate

WF = - Wg

WF = -(-1.55)

WF = 1.55 kJ

(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)

Here the work done by force is not equal to F*d

and it is equal to product of the cos angle and F*d

So, it is not equal to the product of the horizontal displacement and the answer to (a)

During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between two fence posts. The limb exerts a downward force of 176 N on the wire. The left section of the wire makes an angle of 12.5° relative to the horizontal and sustains a tension of 413 N. Find the (a) magnitude and (b) direction (as an angle relative to horizontal) of the tension that the right section of the wire sustains.